The shape behind every quadratic
“
…use a completed square form, e.g. to locate the vertex of the graph of y = ax² + bx + c or to sketch the graph.
The definition, and the words that go with it
A quadratic in is an expression in which the highest power of is . Every one of them can be written
Four pieces of vocabulary, all of which the exam paper will use without explaining:
- , and are the coefficients — is the coefficient of (also called the leading coefficient), is the coefficient of , and is the constant term.
- The condition is what makes it a quadratic at all. If the term disappears and you are left with the straight line . Examiners exploit this: a question with containing an unknown, such as , quietly requires .
- Setting the expression equal to zero gives a quadratic equation, . A value of that satisfies it is called a root, a solution, or a zero — the three words mean the same thing and all three appear in past papers.
- Writing turns it into a curve. Its roots are the -values where that curve crosses the -axis, because crossing the axis is exactly what means.
That last sentence is the hinge of the whole topic. Every algebraic question about a quadratic has a picture behind it, and if the algebra ever stops making sense, drawing the picture will restart it.
The shape
Whatever the numbers, the curve always draws the same shape: a parabola. A parabola is a single smooth U (or an upside-down U). It has no corners, no breaks and exactly one turning point, and it is symmetric — there is a vertical mirror line through that turning point.
The turning point has a name, the vertex. If the curve opens upwards, the vertex is the lowest point on the curve and is called the minimum point; if the curve opens downwards it is the highest point, the maximum point.
The same shape appears outside mathematics: the path of a jet of water, the cable of a suspension bridge and the cross-section of a satellite dish are all parabolas.
A fountain jet, a suspension-bridge cable and a satellite dish. The dashed curve over each one is a parabola, a curve of the form y = ax² + bx + c.
The parts of a parabola that Paper 1 questions ask about, on y = x² − 6x + 4. The curve is symmetric about the dashed line x = 3, so the vertex sits exactly midway between the two roots.
Where the axis of symmetry comes from
The axis of symmetry of is the line . Do not just memorise it — it falls out of the quadratic formula in one line, and seeing that makes it impossible to forget.
The two roots are
The curve is symmetric, so the mirror line must sit exactly halfway between them. Halfway between two numbers is their average, so add them and halve:
The two square roots cancel — one is added, one is subtracted — leaving
So the axis of symmetry is , and the vertex sits on it. Notice this argument never needed the roots to be real: the formula for the axis works even for a curve that never touches the -axis.
- is the -intercept — set and everything else vanishes.
- The sign of is the direction: opens upwards (a minimum), opens downwards (a maximum).
- The axis of symmetry is , and the vertex sits on it. Equivalently: the vertex is exactly halfway between the roots.
The same quadratic can be written three ways, and choosing the right one is most of the skill in this topic. They are all the same curve — they just have different things printed on the outside.
Form | Looks like | Hands you | Reach for it when |
|---|---|---|---|
General | the -intercept | you are starting out, or you need the discriminant | |
Completed square | vertex , and the range | maximum / minimum, range, sketching, “no real roots” | |
Factorised | the roots | solving, inequalities, sign diagrams |
Three ways of writing one quadratic. Choosing the form that hands you the fact you need is the first step in most questions.
One curve, drawn three times. Each form prints a different fact on the outside and leaves the other two greyed out — which is why “which form do I need here?” is usually the first useful question to ask. Nothing about the parabola changes; only what you can read off without working.
When a is negative
If the parabola opens downwards, so the vertex is a maximum and the curve is above the axis between the roots — the opposite of the parabola drawn above. Papers often write such curves as , and it is easy to read the first and miss that . The figure below puts the two side by side.
Both curves are (x − 1)(x − 5); the right-hand one has been multiplied by −1. The roots do not move, because a product is zero the moment a bracket is zero — whatever sits in front of it. Everything that depends on which way the curve opens does move: the minimum becomes a maximum, the range flips, and the region where y is negative jumps from between the roots to outside them.
A clean demonstration — sketching a parabola from nothing
Nothing below needs a calculator or a table of values. Take
Step 1 — which way up? The coefficient of is , which is positive, so the curve opens upwards and its vertex is a minimum.
Step 2 — where does it cross the -axis? Put :
so the curve passes through . This is just , as promised.
Step 3 — where does it cross the -axis? Put and solve:
Factorise by looking for two numbers that multiply to and add to . Those numbers are and :
A product of two things is zero only if one of them is zero, so or , giving
The curve crosses the -axis at and .
Step 4 — where is the vertex? It sits midway between the roots:
Check that against the formula: . ✓
Now substitute back into the equation to get the -coordinate:
So the vertex is , and it is a minimum.
Step 5 — read off the range. The lowest the curve ever gets is , and it climbs forever in both directions, so every -value from upwards is achieved:
Five facts, no plotting. That is a complete sketch.
The same thing with a negative
Now take , written the awkward way round on purpose because that is how papers write it.
Which way up? The coefficient of is . Negative, so the curve opens downwards and the vertex is a maximum. Read the term, not the number at the front.
-intercept: put to get .
Roots: set . Multiply every term by so the term is positive — that makes factorising far easier:
Two numbers multiplying to and adding to : they are and .
Vertex: midway between and is . Then
so the maximum point is .
Range: the curve never goes above , so .
Compare the two demonstrations line by line. The method is identical; only the direction of the two inequality signs changed.
Quadratics carry more marks on Paper 1 than any other topic, and a lot of coordinate geometry turns into a quadratic too. The questions are often among the harder ones, because the quadratic is usually the end of a longer question rather than the whole of it.
Your turn
Nothing here is examined on its own, but everything later in this note assumes you can do it instantly. Do all three without a calculator.
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For the curve , find
(a) the -intercept,
(b) the roots,
(c) the axis of symmetry,
(d) the coordinates of the vertex, and say whether it is a maximum or a minimum,
(e) the range of .Show solution
- 1
(a) Put : so the -intercept is .
The y-intercept is always just c. There is nothing to work out.
- 2
(b) Put :
- 3
Look for two numbers that multiply to and add to . Since the product is positive and the sum is negative, both must be negative: and .
Reading the signs off first — both negative — halves the number of pairs you have to try.
- 4
So or , giving the roots and .
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(c) The axis of symmetry is midway between the roots: (Check: ✓)
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(d) Substitute into the equation: so the vertex is . Since the curve opens upwards, so it is a minimum.
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(e) The curve never goes below its minimum, so the range is .
Answer(a) (b) and (c) (d) minimum at (e)
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- 2
A curve has equation . Find the coordinates of its vertex and state the range of .
Stuck? Show hint
The coefficient of is , not . Decide which way the curve opens before you do anything else, because that decides which way round the final inequality goes.
Show solution
- 1
Identify the coefficients by matching against . Rewriting in the usual order, , so
Rewriting in descending powers first is worth the ten seconds. Reading b off a jumbled expression is where sign errors start.
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is negative, so the curve opens downwards and the vertex is a maximum.
- 3
Axis of symmetry:
Two minus signs. Write the substitution out in full rather than doing it in your head — this is the single most common place to drop a sign.
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Substitute into the original equation:
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So the vertex is , and it is a maximum.
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Since the curve never rises above its maximum, the range is .
Downward parabola ⇒ ⩽, not ⩾. This flip is the whole point of the question.
AnswerMaximum at ; range
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- 3
A parabola crosses the -axis at and , and passes through the point .
(a) Find its equation in the form .
(b) Find the coordinates of its vertex.Stuck? Show hint
If you know the roots, start from the factorised form and use the third point to pin down .
Show solution
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(a) Roots at and mean the brackets and must both be factors, so for some constant .
The bracket for a root at −3 is (x + 3), not (x − 3): the bracket has to be zero when x = −3. Getting this sign backwards is extremely common.
- 2
The constant is not determined by the roots alone — every vertical stretch of the curve has the same roots. Use the third piece of information: the curve passes through , so put and :
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Work out the two brackets:
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Divide both sides by :
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So . Now expand. First the two brackets:
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Then multiply every term by :
Multiply through by the 2 at the very end — carrying it inside the expansion is where the arithmetic goes wrong.
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(b) The vertex is midway between the roots:
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Substitute into whichever form is easier. The factorised form is easier here:
Choosing the friendlier of two equivalent forms is a habit worth building — it is the same idea that makes completing the square so useful.
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So the vertex is .
Answer(a) (b)
- 1
Completing the square
“
carry out the process of completing the square for a quadratic polynomial ax² + bx + c and use a completed square form.
The problem completing the square solves
Try to solve by rearranging. Subtract : . Now what? You cannot divide by without losing a solution, and you cannot square-root the left-hand side because is not a square. The obstacle is that appears twice, so there is nowhere to move it to.
Completing the square removes that obstacle. It rewrites the quadratic so that appears exactly once, inside a bracket that is squared:
The sign means the two sides are equal for every value of — this is a rewriting, not an equation to be solved. And with in one place only, everything opens up: you can undo the square to solve, you can read the vertex off, you can state the range, and you can argue that some value is impossible.
Building up from the perfect square
Before completing a square you have to know what a completed one looks like. Expand patiently:
Stare at the middle term. It is — twice , times . So:
In a perfect square, the coefficient of is always twice the number in the bracket.
Turn that round and you have the whole method. If you are handed and you want it to be a perfect square, then , so — you halve the coefficient of . The bracket must be .
But , and the original had no in it. You have accidentally added , so you must immediately take it away again:
Check by expanding the right-hand side: . ✓
That single line is the whole of completing the square. Everything that follows is bookkeeping around it.
x² + 6x is a square of side x with two strips of width 3 glued on. Those pieces almost form a square of side x + 3 — they are short by exactly the 3 × 3 corner. So adding 9 completes the square, and you subtract the same 9 to keep the value unchanged.
Demonstration 1 — the easiest case,
Express in completed-square form.
Step 1. Deal with the and terms first and leave the constant alone for now:
Step 2. Halve the coefficient of . Half of is , so the bracket is .
Step 3. Write down and correct for what it added. , which is too big, so subtract :
Step 4. Put the original constant back on the end:
Step 5. Collect the two numbers: .
Check. Put into both sides. Left: . Right: . ✓ Do this check every single time — it takes five seconds and catches nearly every arithmetic slip.
Demonstration 2 — a negative , and an odd
Negative . Express in the form .
Half of is , so the bracket is . Then , which is too big:
Note that the number you subtract, , is positive even though was negative — you are subtracting , and a square is never negative.
Odd . Express in completed-square form. Half of is , which is not an integer, and that is completely fine — most exam answers here are fractions.
so
To add and , write as :
Resist the urge to round to unless the question allows decimals — an exam answer of is always safe.
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Take out of the and terms only. Leave the constant outside the bracket.
The constant is not part of the square, so dragging it inside is the single most common source of errors here.
- 2
Halve the coefficient of inside the bracket. That halved number is what goes in the bracket. Half of is , so the bracket is .
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Subtract the square of that number, inside the bracket. expands to , which is too big.
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Multiply back out and collect the constants.
The correction gets multiplied by a — here −4 becomes −36. Forgetting this is the classic slip.
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Check. Expand mentally, or put into both forms: , matching the original constant. Ten seconds, and it catches almost every arithmetic error.
Demonstration 3 — when , in slow motion
Express in the form .
Step 1 — take the factor out of the first two terms only. The constant stays outside the bracket:
Check by expanding back: ✓ and ✓.
Step 2 — complete the square inside the bracket, ignoring everything outside it. Inside we have , which we already know is :
Step 3 — multiply the back in. It hits both things inside the square bracket:
This is the step everyone loses marks on. The was inside the bracket, so it becomes , not .
Step 4 — collect the constants: .
Check with : left gives ; right gives . ✓
Demonstration 4 — when is negative
This is the version examiners like most, because the sign errors multiply. Express in the form .
Step 1 — write it in descending powers of so you can see what you have:
Step 2 — take out the factor from the and terms. Both signs inside the bracket flip:
Expand back to check: ✓ and ✓.
Step 3 — complete the square inside. Half of is , and , so :
Step 4 — multiply the back in. Both terms change sign:
Step 5 — collect: .
Check with : left is ; right is . ✓ And with : left is ; right is . ✓
Because the bracket is subtracted, the largest this expression can ever be is — reached when the bracket is zero, at . So this parabola has a maximum at .
Match the form the question printed
Papers ask for the same rewriting under half a dozen different labels: , , , , , , and even . They are all the same manipulation. What changes is only which letter names which number, and the marks are for the letters the question asked for.
Two traps worth naming:
- If the form is and you obtain , then , not , because the printed form has a inside the bracket.
- If the form is with , , positive integers, and you obtain , then — the minus sign is already in the printed form, so is not .
The variant
Occasionally a question asks for the bracket to keep the -coefficient inside it: "Express in the form ." Do not take the out — the form is telling you not to.
Instead, work backwards from the target. Expanding the target gives
Compare with term by term:
- terms: on both sides. ✓ Nothing to do — this is why the form works.
- terms: , so .
- constants: . With that is , so .
Check with : left is , right is . ✓
Comparing coefficients like this always works and is often quicker than taking the factor out. It is worth having in your toolkit for any "express in the form…" instruction whose shape looks unusual.
The general result
Worth understanding, not memorising — in the exam, do the four steps on the actual numbers.
Why the completed square is worth so much
Written as , the quadratic is telling you the whole geometry of the curve, and the argument is short enough to reconstruct in the exam.
The key fact: a square is never negative. Whatever real number you put in, , and it equals only when the bracket itself is zero, that is when .
Now suppose . Then too, so
for every . The value is therefore a floor that can never go below — and the floor really is reached, at . So is the minimum value, it happens at , and the vertex is .
If the multiplication by reverses the inequality: , so , and now is a maximum.
Completed-square form says exactly how the basic curve y = x² was moved: (x − 2)² − 3 is “2 to the right, 3 down”, so the vertex is (2, −3). The signs are opposite in x and the same in y — a trap worth rehearsing.
For with :
- the vertex is and it is a minimum
- the minimum value of is , occurring at
- the range is
- there are no real roots if (the whole curve sits above the axis)
If , every one of those flips: maximum, , no roots if .
Showing that an expression is always positive
The same argument answers questions such as "show that is positive for all values of ". Complete the square:
Now argue in words. for every real , because a square is never negative. Adding to both sides gives
so is always positive — in fact it is never less than . The final sentence is part of the answer: a completed square on its own is not a proof until you say why it cannot be negative. Later in the course the same move shows that a derivative is always positive, and so that a function is increasing.
Reading the range off — and writing it down properly
The range of a function is the set of values the output can take. For a quadratic that is exactly the argument above, so the completed square hands you the range for free.
On clean numbers. Let for . The bracket is never negative, so the smallest can be is , and it gets there at . The range is
Now . Here the square is subtracted, so the largest can be is , reached at . The range is
Notation matters and is marked. A range is a statement about the output, so it must be written with , or , or — never with . On a 2025 paper, where the range was , the mark scheme states: " scores B0." You can do all the algebra correctly and still lose the mark by naming the wrong letter.
Acceptable: , , "the range is ". Not acceptable: .
The range lives on the y-axis, which is the whole reason it must be written in terms of f(x) or y. The completed square puts the boundary value q in front of you; the sign of a decides whether the range runs upwards or downwards from it.
The archetype: complete the square, then get paid for it twice
(a) Express in the form , where , and are constants. [2]
(b) Hence find the set of values of the constant for which the equation has no real roots. [1]
(c) Find the exact roots of the equation . [2]
Show full working
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(a) Take the out of the and terms only, leaving the outside:
Check immediately by expanding back: 9 × x² = 9x² ✓ and 9 × (−4x) = −36x ✓. Ten seconds spent here saves the whole question.
- 2
Complete the square inside the bracket. Half of is , so the bracket is . Since is too big,
- 3
Substitute that back in, keeping the square brackets so the is clearly outside both terms:
- 4
Multiply the back in. It multiplies the as well as the square:
This is the whole difficulty of the a ≠ 1 case. The −4 lived inside the bracket, so it becomes −36.
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Collect the constants: .
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Check with : the original gives ; the new form gives . ✓
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Now read off the letters against the form the question printed, which was :
The printed bracket is (x + q). Ours is (x − 2), i.e. (x + (−2)), so q = −2. Writing q = 2 is the single most common way to throw away this mark.
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(b) Think about what the equation means graphically: it asks where the curve meets the horizontal line .
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From part (a), the curve's minimum value is , so the curve occupies and nothing below that.
The coefficient 9 is positive, so the parabola opens upwards and −28 is a floor, not a ceiling.
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A horizontal line below that floor cannot touch the curve anywhere, so there are no real roots exactly when
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Note the sign is strict. At the line passes exactly through the vertex and there is a (repeated) root, so must be excluded.
The equal case is a real root, not the absence of one. This is the difference between < and ⩽, and it is worth the mark.
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(c) The word exact forbids decimals, so do not reach for a calculator. Use the completed square from (a), because it has in only one place:
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Add to both sides:
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Divide both sides by :
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Take the square root of both sides. A positive number has two square roots, so the is compulsory:
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Simplify the surd. The square root of a fraction is the square root of the top over the square root of the bottom, and :
√13 does not simplify further — 13 is prime, so it has no square factors. Leaving it as a surd is the point of the word 'exact'.
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Add to both sides:
Leave it here. Writing 3.20 and 0.798 loses the mark — 'exact' means surd or fraction form.
(a) , i.e. , , (b) (c)
Part (b) is a 1-mark question that is impossible without part (a) and trivial with it. That is the entire design of this topic: the completed square is set up in part (a) and then charged for two or three more times. Never abandon a "hence".
Negative leading coefficient, and a form with the minus built in
Express in the form , where and are constants.
Show full working
- 1
First rewrite in descending powers of so nothing is hiding:
Papers write these back to front on purpose. Reordering costs nothing and stops you reading a = 1 off the leading 1.
- 2
Look at the target form, . The squared bracket has no number in front of it and carries a minus sign, so the coefficient of on the right is — which matches. So take the factor out of the and terms:
- 3
Check by expanding back: ✓ and ✓.
Both signs inside the bracket flip. Forgetting to flip the 6x is the most frequent error in this whole sub-topic.
- 4
Complete the square inside the bracket. Half of is , and , so
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Substitute that in:
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Multiply the back in — through both terms in the bracket:
The −9 becomes +9. A minus sign outside a bracket changes every sign inside it, not just the first.
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Collect: .
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Compare with :
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Check with : the original gives ; the new form gives . ✓ And with : original ; new form . ✓
, i.e. and
Because the square is subtracted, this curve has a maximum of at , and its range is . Any question that hands you the form is setting up a maximum, and the next part will almost certainly ask for it.
A form that tells you NOT to take the factor out
Express in the form , where and are constants.
Show full working
- 1
Read the target form carefully. The bracket is , not — the stays inside. So the usual "take the out" move is not what is wanted here.
Matching the printed form is worth both marks. A perfectly correct 4(x − 3/2)² + 4 answers a different question.
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Expand the target so you can compare it with what you have:
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Now compare with , term by term. The terms: on both sides, so there is nothing to fix — which is exactly why this form was chosen.
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The terms: , so dividing by ,
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The constant terms: . Substituting gives
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Subtract from both sides:
Note a² = (−3)² = +9, not −9. Squaring destroys the sign, and that is where this question catches people.
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So Check with : left is , right is . ✓
, i.e. and
Comparing coefficients — expand the target form, match it term by term with what you have, and solve the little equations that fall out — works for any unfamiliar "express in the form…" instruction. It is slower than the standard route on ordinary questions but it never needs you to guess what the examiner meant.
Completed square straight into a range
The function is defined by for .
Express in the form , where , and are constants, and state the range of .
Show full working
- 1
Reorder into descending powers:
- 2
The target form is , in which the coefficient of is . So take the factor out of the first two terms:
Divide each of −2x² and 6x by −2: you get x² and −3x. The sign of the middle term flips, which is the step to slow down on.
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Complete the square inside. Half of is , so the bracket is , and
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So is short of that square, i.e.
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Substitute in:
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Multiply the through both terms:
−2 × (−9/4) = +18/4. Two negatives make a positive; the constant grows rather than shrinking.
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Simplify to , then add :
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So which matches with , , .
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Check with : original gives ; new form gives . ✓
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The range. The square is never negative, so is never positive. The most can be is therefore , reached when the bracket is zero at :
The mark scheme insists on ⩽ and will not accept <. The maximum value is actually attained, so it belongs in the range.
; range
Whenever a Paper 1 question defines a function with a quadratic rule and then asks for a range, complete the square. It is not one method among several; it is the method, and it converts a three-mark question into a one-line read-off.
Completing the square with a letter in it
The function is defined by for , where is a constant.
Given that the range of is , find the possible values of .
Show full working
- 1
The range of an upward parabola is , so the given range is telling us that the minimum value of is . To find that minimum, complete the square — treating as though it were an ordinary number.
Naming what the information means before touching algebra is worth doing out loud. 'Range ⩾ −33' and 'minimum = −33' are the same statement.
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The coefficient of is , so there is nothing to take out. Split off the constant:
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Halve the coefficient of . That coefficient is , and half of is , so the bracket is .
Halving 4a gives 2a — halve the number and leave the letter alone. Students often write 2a² or 4a/2 unsimplified and lose track.
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Expand to see the correction: so it overshoots by , and
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Put the back on:
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The square is never negative, so the minimum value of is the constant part, , reached at .
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Set that equal to :
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This is now an ordinary quadratic equation in . Collect everything on the side that makes the coefficient positive — add and subtract from both sides:
Arranging so the leading coefficient is positive makes the factorisation far easier to spot, and it costs nothing.
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Factorise . The two brackets must multiply to give and ; trying and expanding gives
Trying brackets and expanding to check is fine when the numbers are small. A step-by-step method for factorising when the x² coefficient is not 1 (splitting the middle term) is in the next section; the quadratic formula also works here.
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So , giving or :
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Both are genuine answers. The question said "the possible values" (plural) and nothing in it rules either out, so report both.
The mark scheme is explicit that discarding one of them loses the mark. Only reject a root when the question gives you a reason to.
or
Completing the square works exactly the same way with letters in it — you are halving instead of halving . The pattern "you are told the range / minimum, find the constant" is very common, and it always turns into an equation in that constant, usually itself a quadratic.
The −4 is inside the bracket, so it must be multiplied by the 9 on the way out.
Squaring the bracket introduces an extra +9 that was never in the original — you have to take it back out.
has a minimum at
has a minimum at
The bracket is zero when x = 2. The sign inside the bracket is opposite to the position of the vertex.
Obtaining for the form and writing
The printed form has a plus inside the bracket, so q is whatever must be added. (x − 2) is (x + (−2)).
Stating the range of as
(or )
A range is a set of output values. The mark scheme for exactly this says 'x ⩾ 8 scores B0' — correct working, wrong letter, no mark.
For in the form with , , positive, answering
The minus sign is already printed in the form, so a supplies only the size. The question even tells you a is a positive integer.
Rounding to
Keep exact fractions unless the question allows decimals
5.3 is not equal to 21/4, so the identity is no longer true. Fractions are not 'unfinished' — they are the answer.
Completing the square is usually a short part (a), worth 2 or 3 marks, and the wording barely changes. What does change is the printed form — recent papers have used , , , , , and . The later parts then use your answer for a range, a minimum point, a "no real roots" argument or an exact solution, so an error here costs marks further on too.
Your turn
Four completed squares, deliberately in four different printed forms. Check every one by substituting x = 0 into both sides before you look at the solution.
- 19709/12 M/J 2025 Q11(a)2 marks
Express in the form , where and are integers.
Show solution
- 1
The coefficient of is , so nothing needs taking out. Separate the constant:
- 2
Halve the coefficient of : half of is , so the bracket is .
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, which is more than , so subtract the back off:
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Put the original back:
- 5
Collect: .
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Check with : left ; right . ✓ So and .
Answer, i.e. ,
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- 29709/11 M/J 2024 Q1(a)2 marks
Express in the form , where and are constants.
Stuck? Show hint
The variable is rather than , which changes nothing at all. Take the out of the first two terms only.
Show solution
- 1
Take out of the and terms, leaving outside:
Check: 3 × y² = 3y² ✓ and 3 × (−4y) = −12y ✓.
- 2
Complete the square inside. Half of is , and , so
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Substitute in:
- 4
Multiply the through both terms:
The −4 inside becomes −12. This is the step the marks are really for.
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Collect: .
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Check with : left ; right . ✓ So , .
Answer, i.e. ,
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- 39709/15 O/N 2025 Q33 marks
(a) Express in the form , where , and are rational constants to be determined. [2]
(b) The curve with equation and the line have exactly one point of intersection. Using your answer to part (a) or otherwise, state the value of the constant . [1]
Stuck? Show hint
For (b): a horizontal line meets an upward parabola exactly once only when it passes through one very particular point.
Show solution
- 1
(a) Take out of the and terms:
10 ÷ 4 = 5/2. Simplify the fraction now, before halving it, or the arithmetic gets unpleasant.
- 2
Halve : half of is , so the bracket is .
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, so
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Substitute in:
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Multiply the through. :
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Write as and collect: .
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Check with : left ; right . ✓
- 8
(b) The curve opens upwards (the is positive) with minimum value . A horizontal line above the minimum cuts the curve twice; below it, not at all. Exactly one intersection happens only when the line passes through the vertex itself.
Sketch the three cases if this is not obvious — a horizontal line sliding up through a U-shape. Only one height gives a single touch.
- 9
So equals the minimum value:
- 10
Write it as , not . The mark scheme rejects the latter.
k is a y-value: it is the height of the horizontal line. Naming the wrong variable throws the mark away.
Answer(a) (b)
- 1
- 49709/12 M/J 2023 Q33 marks
(a) Express in the form , where and are integers and is to be given in terms of the constant . [2]
(b) Hence or otherwise find the set of values of for which the equation has no real roots. [1]
Stuck? Show hint
Treat exactly like any other number — it is only ever carried along, never operated on. For (b), where must the whole curve sit if it never crosses the -axis?
Show solution
- 1
(a) Take out of the and terms only. The is the constant term, so it stays outside:
Dragging p inside the bracket is the classic error here, and it wrecks both marks.
- 2
Complete the square inside. Half of is , and , so
- 3
Substitute in:
- 4
Multiply the through:
- 5
There is nothing further to collect, since and are unlike terms. So giving , , .
- 6
(b) The coefficient of the squared bracket is , so the curve opens upwards and its minimum value is .
- 7
"No real roots" means the curve never reaches the -axis. For an upward parabola that means the whole curve sits strictly above the axis, i.e. its minimum is positive:
Strictly above. If the minimum were exactly 0 the curve would touch the axis and there would be a repeated root — which is a real root.
- 8
Add to both sides:
Answer(a) (b)
- 1
Solving a quadratic equation
“
solve quadratic equations, and quadratic inequalities, in one unknown.
What "solve" means, and why there are three methods
To solve is to find every value of that makes the statement true. Graphically, you are finding where the curve crosses the -axis, so there can be two answers, one, or none at all.
This is the skill you will use most often in the whole topic. You are rarely handed a quadratic and asked to solve it. Usually the quadratic results from something else — the end of a coordinate-geometry question, a series question or a trigonometry question — and you have to finish the job under time pressure.
You have three tools. They all give the same answers; they differ only in speed and in what form the answer comes out in.
Factorising — fastest, but only when the roots are whole numbers or simple fractions
The formula — always works; the default when factorising fails
Completing the square — when asked for by name, or when the question wants exact surds
Method 1 — factorising, and the fact that makes it work
Factorising rewrites the quadratic as a product of two brackets. The reason that helps is a fact about numbers so simple it usually goes unsaid:
If two numbers multiply to give zero, at least one of them is zero.
Nothing else has this property. If you learn almost nothing about and ; but if then or , with no third possibility. So once you have
you can immediately split it into two linear equations, and , and read off and .
This is why you must always move everything to one side first. From you can conclude nothing at all, because plenty of pairs of numbers multiply to . Only zero on the right-hand side lets you split the brackets.
Factorising when
For , you need two numbers that multiply to and add to . Then those two numbers go straight into the brackets.
Why? Because expanding gives
so the coefficient of is the sum and the constant is the product. Read that backwards and you have the rule.
The signs tell you where to look, which cuts the search enormously:
- positive, positive → both numbers positive.
- positive, negative → both numbers negative.
- negative → one positive and one negative, and the bigger one carries the sign of .
On clean numbers. Solve . Product , sum : the product is positive and the sum is negative, so both numbers are negative. Pairs multiplying to : , , . The one whose (negative) sum is is and :
Another. Solve . Product , so one number is negative. Sum , so the positive one is bigger. Pairs: and , and . Taking and gives sum ✓:
Factorising when — splitting the middle term
For the trick still works, but with one extra step. You look for two numbers that
- multiply to , and
- add to ,
then use them to split the middle term into two, and factorise in two halves.
On clean numbers. Solve .
Step 1. , and . So find two numbers multiplying to and adding to . Since the product is negative, one is negative. Pairs multiplying to : , , , , . The pair and differ by , so take and :
Step 2. Split the into . Nothing has changed in value:
Step 3. Factorise the first two terms and the last two terms separately. From take out ; from take out :
Step 4. The bracket is now a common factor of both pieces, so pull it out:
If the two brackets in step 3 had not matched, you would know a sign had gone wrong somewhere — that agreement is a built-in check.
Step 5. Set each factor to zero:
Check one of them: putting into the original gives ✓
Two special cases that trip people up
No constant term. has a common factor of :
Do not divide both sides by to get . That throws away the solution , and mark schemes catch it constantly. Factorise the out and keep it as a root; if the context then rules out (a number of terms, a length), reject it afterwards, explicitly.
No middle term. needs no search at all. Either recognise the difference of two squares, , or rearrange to and take both square roots: . Writing only loses half the answer, and the same mistake later spoils the disguised quadratics in the section "Equations that are quadratic in something else".
Method 2 — the formula, and where it comes from
Most quadratics do not factorise over the integers, and then you need the formula. It is on the formula list, but deriving it once is worth the five minutes: the derivation is nothing but completing the square on the general quadratic, and it explains why (the discriminant, later in this note) matters so much.
Start from with .
Divide every term by , so the leading coefficient becomes :
Move the constant across:
Complete the square on the left. Half of is , so the bracket is , and squaring it adds . Add that to both sides to keep the equation balanced:
Put the right-hand side over the common denominator . Note :
Square-root both sides, remembering the , and using :
Subtract and put the two fractions over one denominator:
Every feature of the formula is now explained: the came from doubling in the completing-the-square step, the came from the bracket, and the thing under the root came from combining two fractions. If that expression is negative there is no real square root, and therefore no real solution — the idea behind "The discriminant" section.
Using the formula on clean numbers
Solve , giving exact answers and then answers to 3 significant figures.
Step 1 — write down , , explicitly, with their signs:
Step 2 — substitute, keeping every bracket. This is not fussiness; brackets are what stop squared becoming :
Step 3 — evaluate the pieces one at a time.
Step 4 — stop here if the question says exact. has no square factors, so cannot be simplified.
Step 5 — only if decimals are wanted, , so
Sanity check. The two roots should average to . And indeed ✓
Method 3 — completing the square, for exact answers
You already know how to complete the square from the "Completing the square" section. To turn it into a solution, undo the square:
Do not memorise that chain — do the four moves on the actual numbers. What matters is the shape of the answer: a number, plus-or-minus a surd. Whenever a question asks for solutions "in the form ", or says "exact", it is telling you that the answer looks like this and that a decimal will score nothing.
“Exact” is an instruction, not a suggestion
If a question says exact, decimals score zero. Leave your answer as a surd or a fraction: , not or .
Conversely, if a question sets an interval in degrees and asks for angles, it wants decimals, normally to 1 d.p. And "give your answer correct to 3 significant figures" means exactly that — an exact surd left unsimplified there can also lose the mark.
Read the last line of the question before you start writing, and decide then which kind of answer you are heading for.
The question says… | Use | Because |
|---|---|---|
“Solve”, with small whole-number coefficients | Factorising | fastest, and no rounding to get wrong |
“Find the exact solutions / exact roots” | Completing the square, or the formula left as a surd | the answer is a number ± a surd |
“Use completing the square to…” | Completing the square | naming the method makes it compulsory |
“Use the quadratic formula to show that…” | The formula | same — the method is the thing being marked |
“Giving your answers in the form ” | Completing the square | the printed form is a hint about the shape of the answer |
“Correct to 3 significant figures” | The formula, then round at the very end | keep full accuracy until the last line |
The question almost always tells you which tool it wants. Naming a method in the question makes that method compulsory — the marks are for the working, not the numbers.
Solving exactly by completing the square
Use completing the square to find the exact solutions of the equation .
Show full working
- 1
The question names the method, so completing the square is compulsory here — the formula would get the right numbers but score only 1 of the 2 marks.
"Use completing the square to…" is an instruction about the working. The mark scheme awards the method mark only for producing an (ax + b)² term, and gives just one special-case mark for correct answers from another method.
- 2
Take the out of the and terms, leaving the outside:
Check: 4 × x² = 4x² ✓ and 4 × (−x) = −4x ✓. The coefficient of x inside is −1, which is easy to misread as −4.
- 3
Complete the square inside the bracket. Half of is , so the bracket is , and
- 4
That overshoots by , so
- 5
Substitute back in:
- 6
Multiply the through both terms. Note :
- 7
Collect the constants:
- 8
Now undo the square, one move per line. Add to both sides:
- 9
Divide both sides by and simplify the fraction:
- 10
Take the square root of both sides, with :
Both signs. Half the marks in exact-solution questions are lost by taking only the positive root.
- 11
Add to both sides:
- 12
That is already an acceptable exact answer. To tidy it, note , so
All of these forms are accepted. Tidying is optional here — but part (b) of this question reuses the answer, and the tidy form is much easier to work with.
, i.e.
Look at the shape of the answer: a number a surd, both over the same denominator. That is what "exact" always produces. If your working ever hands you and in a question that said exact, you have used the wrong tool.
Forming the equation in the first place
In a real Paper 1 question the quadratic is rarely printed. You have to build it, and the mark scheme usually awards a mark for "simplify to a three-term quadratic" before any solving happens at all. The recipe is always the same:
- Write down what the question tells you as equations, with letters for what you do not know.
- Eliminate variables until one equation contains one unknown.
- Expand every bracket and clear every fraction.
- Collect all terms on one side so the right-hand side is , with a positive coefficient if you can arrange it.
- Solve — and then check the answers against the context, because a length, a number of terms or a common ratio may make one of them impossible.
The two worked examples below are that recipe in two very different disguises.
Forming a quadratic from a geometric condition
Three points , and have coordinates , and , where is a constant. It is given that the angle is a right angle.
Show that one of the possible values of is , and find the other possible value.
Show full working
- 1
Angle is the angle at , between the lines and . Two lines are perpendicular exactly when the product of their gradients is , so that is the condition to write down.
Translate the geometry into an algebraic condition before doing any arithmetic. 'Right angle at R' means gradient RP × gradient RQ = −1, nothing else.
- 2
Gradient of , using with and :
2 − (−13) = 2 + 13 = 15. Write the double negative out; it is a favourite place to drop a sign.
- 3
Gradient of , using and :
- 4
Multiply them and set the product equal to :
- 5
Combine the two fractions into one by multiplying tops and bottoms:
- 6
Multiply both sides by to clear the fraction:
- 7
Expand the left-hand side term by term:
Four products, written out. Rushing this expansion is where this question is usually lost.
- 8
So
- 9
Collect everything on the side that makes positive. Add , subtract , add to both sides:
- 10
Simplify the constants: .
This three-term quadratic is worth a mark on its own in the mark scheme, before any solving.
- 11
Factorise. Two numbers multiplying to and adding to : since the product is negative one is negative, and the negative one must be bigger. Try and : product ✓, sum ✓.
- 12
So or :
- 13
The question asked us to show that is a possible value, and we have derived it rather than merely checking it — so the "show that" is complete — and the other value is .
In a 'show that … and find the other', deriving both from the same quadratic answers both halves at once. Verifying k = 10 by substitution only would score a special-case single mark.
or
Perpendicular lines, distances between points, and areas of triangles are the three commonest ways coordinate geometry hands you a quadratic. In every case the routine is identical: write the geometric condition as an equation, expand it, collect to zero, factorise.
When the roots themselves are the unknowns
A function is defined by for , where and are constants.
It is given that and the roots of are and , where is a constant. Find the values of and .
Show full working
- 1
Write down what we have. With , and we are told its two roots are and .
- 2
Here is the key idea. If a quadratic has roots and , then and are factors, so the quadratic must be for some constant — and that has to be the leading coefficient, because expanding gives
This is the factorised form from the first section, used in reverse. Knowing the roots determines the quadratic up to the multiplier out front.
- 3
Our roots are and , so
- 4
Expand the two brackets carefully:
- 5
Collect the middle terms: .
- 6
Multiply through by :
- 7
Now compare coefficients with . Two quadratics are identical only if every matching coefficient is equal.
- 8
Coefficient of : . True automatically — no information, which is expected since we built it that way.
- 9
Coefficient of :
- 10
Constant term: Divide both sides by :
- 11
Two equations, two unknowns. Rather than substituting, notice that contains , and tells us . Substituting that into :
Spotting that pm² factors as (pm) × m turns a messy simultaneous pair into a one-line answer. Always look for the block you already know.
- 12
So
- 13
Substitute back into , :
- 14
Check. With and the function is , and the claimed roots are and . Test : ✓. Test : ✓
On a five-mark question with two unknowns, a substitution check is cheap insurance and takes under a minute.
and
"The roots are and " is an invitation to write the quadratic as and compare coefficients. It is much less algebra than substituting each root into the equation and solving the pair simultaneously — which is the other route the mark scheme allows, and takes twice as long.
or
Expand, collect to zero, then factorise:
Splitting a product only works when the product equals zero. Nine is not zero, and dozens of number pairs multiply to give it.
or
Dividing both sides by x silently deletes the root x = 0. Factorise it out instead, then reject it only if the context genuinely forbids it.
In the formula, writing when
Substitute with brackets round every negative. A squared quantity is never negative, so a negative b² is always an error.
Every positive number has two square roots. Losing the negative one costs a mark here and wrecks disguised quadratics later.
Giving and when the question said “exact”
‘Exact’ forbids decimals outright. Stop the moment you have the surd.
In recent papers, a quadratic to solve has appeared at the end of questions on arithmetic progressions, binomial coefficients, perpendicular gradients, volumes of revolution, composite functions and circles. Being fast and reliable at this one skill protects marks across the whole paper.
Your turn
The first two are pure technique; the last two are the kind of forming-and-solving that Paper 1 actually asks for. Do the first without a calculator.
- 1
Solve
(a)
(b)
(c)Show solution
- 1
(a) Two numbers multiplying to and adding to . The product is positive and the sum negative, so both are negative: and .
- 2
So or , giving or .
- 3
(b) Here , so use the split-the-middle method. , and . Two numbers multiplying to and adding to : and .
- 4
Split the into :
The order does not matter — 2x² − x + 6x − 3 factorises just as well — but keeping the larger piece first usually makes the common factor easier to see.
- 5
Factorise in pairs. From take out ; from take out :
- 6
Both brackets agree, so pull out :
- 7
So or , giving .
- 8
(c) There is no constant term, so there is a common factor of :
Do not divide through by x. Factorising it out keeps x = 0 as a genuine root.
- 9
So or , giving or .
Answer(a) (b) (c)
- 1
- 2
Solve , giving your answers (a) exactly, and (b) correct to 3 significant figures.
Stuck? Show hint
Try to factorise first — you will not manage it, and that is the signal to use the formula.
Show solution
- 1
Look for two numbers multiplying to and adding to . The only integer pairs are (sum ) and (sum ). Neither works, so it does not factorise over the integers — use the formula.
Spending ten seconds ruling factorisation out is worth it. Hunting for factors that do not exist is how minutes disappear.
- 2
Write down the coefficients with their signs:
- 3
Substitute into , bracketing every negative:
- 4
Evaluate the pieces: ; ; , so ; and .
c is negative, so −4ac is positive and the number under the root grows. Getting this sign wrong gives √13 instead of √37.
- 5
- 6
(a) is prime, so the surd cannot be simplified. The exact answers are
- 7
(b) , so
- 8
To 3 significant figures: or .
−0.180 needs the trailing zero: the three significant figures are 1, 8 and 0. Writing −0.18 is only two.
Answer(a) (b) or
- 1
- 39709/12 O/N 2024 Q2(b)3 marks
The first term of an arithmetic progression is and the common difference is . The sum of the first terms is times the sum of the first terms.
Find the value of .
Stuck? Show hint
Use twice, once with and once with . When you reach the quadratic, do not divide by .
Show solution
- 1
The sum of the first terms of an arithmetic progression is . Substitute and :
You meet this formula properly in the Series note. Here it is only the source of the quadratic.
- 2
Simplify inside the bracket: and , so
Simplifying the inside bracket once, in general n, saves doing the same arithmetic twice.
- 3
With :
- 4
With :
- 5
Expand that one:
- 6
Now write down the condition " is times ":
- 7
Simplify the right-hand side. , so
- 8
Expand:
- 9
Collect everything on the right so the coefficient is positive. Subtract and add to both sides:
- 10
There is no constant term, so factorise out the common factor :
Dividing by k here would lose the root k = 0 — which happens to be the one we discard, but you must show it and discard it, not delete it silently.
- 11
So or .
- 12
Reject : counts terms of a progression, so it must be a positive whole number. Hence
State the reason. 'Reject k = 0 since k is a number of terms' is what turns an extra root into a correct final answer.
Answer - 1
- 4
A rectangle has a perimeter of cm and an area of cm. Find its length and width.
Stuck? Show hint
Call the two sides and . The perimeter gives one equation and the area another; use the first to get rid of one letter.
Show solution
- 1
Let the length be cm and the width be cm. Perimeter is twice the length plus twice the width:
- 2
Divide every term by to make life easier:
- 3
Area is length times width:
- 4
Make the subject of the linear equation :
Always rearrange the linear equation, never the quadratic one. The simultaneous-equations section later in this note explains why.
- 5
Substitute into :
- 6
Expand the left-hand side:
- 7
Collect everything on the side that makes positive. Add and subtract from both sides:
- 8
Factorise: two numbers multiplying to and adding to , both therefore negative. and work.
- 9
So or .
- 10
Find the partner for each from : if then ; if then .
- 11
Both roots are positive, so neither is impossible — they are the same rectangle described twice. The sides are cm and cm.
Here the 'second root' is not spurious, it is the same answer with the labels swapped. Check what a root means before rejecting it.
AnswerThe rectangle is cm by cm
- 1
Quadratic inequalities
“
solve quadratic equations, and quadratic inequalities, in one unknown.
The genuinely new skill
At IGCSE or O Level you solved quadratic equations. The A Level addition is the inequality — questions phrased as "find the set of values of for which…", "determine the set of values of …", "find the range of possible values of ". This is where marks leak, for two reasons:
- Students try to do it by algebra alone, treating as though it behaved like . It does not.
- Even with the right two numbers, the answer gets written down in a notation the mark scheme refuses.
Both problems are fixed by the same habit: find the critical values, then look at a picture, then write the answer as a single object. This section builds that habit from the ground up.
First, the one rule about inequalities
An inequality can be added to, subtracted from, multiplied and divided just like an equation, with exactly one exception:
Multiplying or dividing both sides by a negative number reverses the inequality sign.
Test it on numbers so you believe it. Start from the true statement . Multiply both sides by :
On the number line is to the right of , so . The sign has turned round.
The practical consequence for this section is one piece of advice: arrange the quadratic so the coefficient is positive, by moving terms across rather than by multiplying by . Then you never have to remember the rule.
Why you cannot just split the brackets
It is tempting to factorise and then argue factor by factor, the way you do with an equation. Watch it fail.
Suppose . If you say "so and ", you get and , i.e. . Test : , which is positive, not negative. The "answer" is wrong everywhere.
The reason is that a product is negative when one factor is negative and the other is positive — and there are two ways for that to happen. Chasing both cases by hand is possible but slow and error-prone. Instead, use one of the two reliable methods below. They are the same idea in two dresses: work out the sign of the quadratic on each stretch of the number line.
The critical values
Whichever method you use, the first move is identical. The quadratic can only change sign where it is zero — a continuous curve cannot get from positive to negative without passing through zero. So:
Solve the corresponding equation first. Its roots are called the critical values, and they cut the number line into at most three stretches. On each stretch the sign of the quadratic is constant.
That is the whole theory. Everything else is deciding which stretches you want.
Method A — the sketch (recommended)
Solve , and then .
Step 1 — critical values. Set the quadratic to zero: . Two numbers multiplying to and adding to are and :
Step 2 — sketch. You need almost nothing on this sketch: the coefficient of is , so it is a U opening upwards, and it crosses the axis at and at . Draw that in five seconds. Do not compute the vertex, do not plot points.
Step 3 — read it off. Between and the U dips below the axis, so that is where . Outside and both arms are above the axis, so that is where . Therefore
Step 4 — check with one test point, which costs three seconds and catches a reversed answer. Take , which is in "": ✓. Take , which is between: ✓.
The sketch for x² − 5x + 4. The curve is below the axis exactly between its roots 1 and 4, and above the axis outside them. Dropping the roots down onto a number line turns the sketch directly into the answer.
Method B — the sign table
If you would rather not draw, tabulate. Factorise into and work out the sign of each bracket on each stretch. Pick any convenient number in the stretch — the sign cannot change within it.
- For , try : (negative), (negative). Negative negative positive.
- For , try : (positive), (negative). Positive negative negative.
- For , try : (positive), (positive). Positive.
Written as a table it looks like this:
x < 1 | x = 1 | 1 < x < 4 | x = 4 | x > 4 | |
|---|---|---|---|---|---|
− | 0 | ||||
− | − | − | 0 | ||
+ | 0 | − | 0 | + |
A sign table for x² − 5x + 4. The bottom row is the answer: the quadratic is positive outside the critical values and negative between them — exactly what the sketch showed.
For an upward parabola () with roots :
Less than zero → between. Greater than zero → outside. If , both swap — which is precisely why you should rearrange to make positive first.
- 1
Get everything to one side, so you have or . Keep the coefficient positive if you can.
If you multiply or divide an inequality by a negative number, the sign flips. Avoiding that is easier than remembering it.
- 2
Find the critical values — the roots of the corresponding equation. Factorise if you can, otherwise use the formula or complete the square.
- 3
Sketch the parabola. You only need the two roots and which way it opens; nothing else on the sketch matters.
- 4
Read the answer off the sketch: the parts of the -axis where the curve is on the side you want.
- 5
Decide whether the endpoints are included. Strict or excludes them; or includes them.
At a critical value the quadratic is exactly zero, so it belongs to a ⩽ answer and not to a < answer. Copy the question's signs.
- 6
Write it in one piece. Between the roots is a chain: . Outside is two statements joined by or: or .
- 7
Test one value from your answer set back in the original inequality.
Three seconds, and it catches the single commonest error in the topic — giving the complement of the right answer.
Demonstration — a negative coefficient
Solve .
Step 1 — make the coefficient positive. Rather than multiplying by and having to remember to flip, move every term to the other side. Add and subtract and from both sides:
which reads more naturally the other way round as
Notice the sign turned round automatically — because the terms swapped sides, not because we remembered a rule.
Step 2 — critical values. . Two numbers multiplying to and adding to are and :
Step 3 — sketch. Upward U crossing at and . We want where it is at or below zero, which is between the roots, including the two ends because the sign is .
Step 4 — test. Take : the original expression is ✓. Take : , which is not ✓ (correctly outside our set).
Demonstration — when it does not factorise
Solve .
Step 1 — critical values. It does not factorise over the integers (no pair multiplies to and adds to ), so complete the square or use the formula. Completing the square:
Setting that to zero: , so and
Step 2 — order them. , so the critical values are approximately and . Knowing which is smaller matters for writing the answer, and a rough decimal is the quickest way to be sure.
Step 3 — read it off. Upward parabola, and we want it above zero, so we want the outside:
Leave the surds in unless the question asks for decimals. Notice that this is the one case where the answer genuinely is two separate statements — and it must be joined by or, not written as a chain.
The demonstration's sketch to scale: above zero (green) outside the surd roots 3 − √5 and 3 + √5, below zero (coral) between them. The green pieces drop onto a number line as two separate arrows — hence “or”, never a chain.
The notation that mark schemes actually reject
Examiners write these rules into their mark schemes, so the notation costs marks directly.
A "between" answer must be one chain. For , the published guidance reads: "Condone , and and — but not or ." The word or turns the statement into "every real number", which is not an interval at all.
An "outside" answer must use or. Writing for or is nonsense, and the guidance on one such question says flatly "Do not accept ."
Match the strictness of the question. On a "do not meet" question the guidance reads "A0 if sign or signs used"; on another, "B0 for use of and/or ." Strict in, strict out.
Critical values are not an answer. Finding and and stopping is typically one mark out of two; the directed inequality is the other.
Reading “decreasing” as an inequality
A curve is such that .
Find the set of values of for which decreases as increases.
Show full working
The finished sketch this working builds: critical values −4 and 2/3 dropped onto a number line, curve below zero between them — so the answer reads off as one chain, −4 < x < 2/3.
- 1
is the gradient of the curve at each value of . You learn how to find it in the Differentiation note; here it is given, so all you need is what its sign means.
A positive gradient means the curve goes uphill from left to right; a negative gradient means it goes downhill.
- 2
Translate the words. " decreases as increases" means the curve is going downhill, which means its gradient is negative:
This translation is a mark on its own in the mark scheme. Write the inequality down before doing anything with it.
- 3
Substitute the given gradient:
- 4
The coefficient is already positive and everything is already on one side, so go straight to the critical values. Solve
- 5
Since , split the middle term. , and . Two numbers multiplying to and adding to : and .
- 6
Split the into :
- 7
Factorise in pairs. From take ; from take :
- 8
The brackets match, so pull out :
- 9
So or , giving the critical values
- 10
Sketch: an upward parabola cutting the axis at and . We want it below zero, which for an upward parabola is between the roots.
Say 'below zero ⇒ between' out loud. It is the sentence that decides the mark.
- 11
Write the answer as one chain, with the smaller value first:
- 12
Strict inequalities, because the question is about decreasing — at and the gradient is exactly zero, so the curve is momentarily flat rather than decreasing.
- 13
Test: at , which is inside the interval, ✓. At , which is outside, ✓.
The published guidance for this exact question says: "Condone , … but not or ." One three-letter word is the difference between full marks and losing the last one.
An inequality buried in a differentiation question
A function is such that for .
Determine the set of values of for which is decreasing.
Show full working
- 1
is another way of writing the gradient of . "Decreasing" means the gradient is negative, so the condition is :
Writing this line down is worth a mark on its own — it is the translation from words into algebra.
- 2
The bracket has to come apart before this is a quadratic in standard form. Expand first, on its own:
Expand the square as a separate line. Trying to multiply by 6 at the same time is where the arithmetic goes wrong.
- 3
Now multiply that by :
- 4
Bring in the :
- 5
Collect the two terms: .
- 6
Every coefficient is divisible by . Divide through — and since is positive, the inequality sign does not move:
Simplify before factorising. Splitting the middle term of 24x² − 78x + 54 means hunting for factors of 1296; after dividing it is factors of 36.
- 7
Find the critical values from . Here and ; two numbers multiplying to and adding to are and .
- 8
Split:
- 9
Factorise in pairs — from take , from take :
- 10
Pull out the common bracket:
- 11
So the critical values are and , i.e. .
- 12
Upward parabola, and we want it below zero, so the answer is between the roots. With :
- 13
Test , inside the interval: ✓
This is filed as a differentiation question, and it is really a quadratic inequality wearing a hat. The mark scheme explicitly rejects " or " — that phrasing describes everything, not an interval. Write the chain.
A completed square, square-rooted into a double inequality
(a) Express in the form , where and are constants. [2]
(b) The function is defined by for , where and are constants. The function is defined by for .
Given that it is possible to form the composite function , find the least possible value of and the greatest possible value of . [3]
Show full working
- 1
(a) was worked in full in the "Completing the square" section (comparing coefficients with ). In brief: gives , and gives , so
- 2
(b) means "do first, then feed the result into ". For that to make sense, every output of must be an allowed input of . The allowed inputs of are , so we need
Composite functions are taught in the Functions note; this one fact is all you need here. Getting from the words 'it is possible to form gf' to the inequality f(x) < 8 is the first mark.
- 3
Using the completed square from part (a):
- 4
Subtract from both sides:
- 5
Now think about what this says. A square is less than exactly when the thing being squared lies strictly between and — because both and equal , and anything further out squares to more than .
Writing 2x − 3 < 2 alone loses half the solution. Squaring destroys signs, so undoing it produces a two-sided statement.
- 6
So
- 7
Solve the chain by doing the same thing to all three parts. Add throughout:
- 8
Divide all three parts by . Two is positive, so no signs turn round:
- 9
Compare that with the given domain . The composite exists precisely when the domain of sits inside , so the smallest can be is and the largest can be is :
- 10
Alternative route, if you prefer the standard method: from expand to , so , which factorises as with critical values and — below zero means between, giving the same chain.
The mark scheme accepts both. The square-root route is faster when the completed square is already sitting in front of you from part (a).
(a) (b) least , greatest
becomes , and becomes or . Those two lines turn up constantly in discriminant work (the next section), where you routinely reach things like .
Multiplying an inequality by −1 reverses it. Safer still: move everything to the other side instead of multiplying.
or
“Or” makes it true for every real number. A between-the-roots answer must be written as a single chain.
for an “outside the roots” answer
or
The mark scheme on one such question says simply 'Do not accept 0 < k < 4/3'. Outside means two separate pieces, joined by 'or'.
and
Find the critical values, sketch, then read the interval off
A product being negative does not mean both factors are negative — that reasoning quietly assumes what it is trying to prove.
Giving the answer as
The critical values are the working; the set of values is the answer.
Undoing a square always produces two bounds. Keeping only the positive one throws away half the solution set.
Writing when the question used a strict inequality
Mark schemes say 'A0 if ⩽ sign or signs used'. Copy the strictness of the question exactly.
Quadratic inequalities are almost never set on their own. The usual phrasings are "find the set of values of for which decreases", "find the set of values of for which the line does not meet the circle", "find the range of possible values of ", "find the least possible value of ". The quadratic inequality is the final two lines of a differentiation, coordinate-geometry, functions or integration question. Which is exactly why the notation matters so much: by the time you reach it you are short of time, and the temptation to scribble two inequalities joined by "or" is at its strongest.
Your turn
Sketch every one, even the ones you think you can do in your head. Then check your notation against the question's inequality signs before you move on.
- 1
Solve the inequality .
Show solution
The solution sketch: filled dots at −5 and 2 because the sign is “or equal to” — the critical values themselves satisfy the inequality and belong to the answer.
- 1
Everything is already on one side and the coefficient is positive, so go to the critical values:
- 2
Two numbers multiplying to and adding to . The product is negative so one is negative; the sum is positive so the positive one is bigger. Try and : product ✓, sum ✓.
- 3
So the critical values are and .
- 4
Sketch an upward parabola crossing at and . We want where it is at or above zero, which for an upward parabola is outside the roots.
Above zero ⇒ outside. If you cannot remember it, test x = 0: 0 + 0 − 10 = −10, which is negative — so the middle is the wrong region.
- 5
The sign is , so the critical values themselves are included (the quadratic is zero there, and zero satisfies ).
- 6
Write the answer as two statements joined by or:
- 7
Check: at , ✓; at , , correctly excluded ✓.
Answeror
- 1
- 2
Solve the inequality .
Stuck? Show hint
The coefficient is negative. Move everything across to the other side rather than multiplying by , and let the sign turn round by itself.
Show solution
- 1
Move every term to the right-hand side so that the term becomes positive. Add and to both sides, and subtract :
- 2
Read it the natural way round:
0 < A and A > 0 say the same thing. Rewriting it with the quadratic on the left is just for readability — no rule was used.
- 3
Critical values: solve . Two numbers multiplying to and adding to are and .
- 4
Upward parabola crossing at and ; we want it strictly above zero, so outside the roots, endpoints excluded:
- 5
Check against the original inequality, which is the safest place to test. At : ✓. At : is false, correctly excluded ✓.
Always test in the original. If a sign flipped somewhere, testing in the rearranged version would agree with your mistake.
Answeror
- 1
- 3
Solve the inequality , giving your answer in exact form.
Stuck? Show hint
It does not factorise. Complete the square to get the critical values — and remember to work out roughly how big is so you know which critical value is the smaller.
Show solution
- 1
Try factorising first: two numbers multiplying to and adding to . Since is prime the only integer pair is and , whose sum is or . No good — so complete the square.
- 2
Half of is , and , which is too big:
- 3
Find the critical values by setting that to zero:
- 4
Square-root both sides, with :
- 5
Work out roughly which is which: , so the critical values are about and . The smaller is .
Order matters for writing a chain. A rough decimal takes two seconds and removes all doubt.
- 6
Upward parabola, and we want it at or below zero, so the answer is between the roots, endpoints included because the sign is :
- 7
Check: at (comfortably inside), ✓. At (outside), , correctly excluded ✓.
- 8
Shortcut worth noticing. From the completed square, the inequality is , and undoing the square directly gives , hence the same chain. On a "between" inequality that route is one line shorter.
Answer - 1
The discriminant
“
find the discriminant of a quadratic polynomial ax² + bx + c and use the discriminant, e.g. to determine the number of real roots of the equation ax² + bx + c = 0. Knowledge of the term ‘repeated root’ is included.
Counting the roots without finding them
Sometimes a question does not want the roots. It wants to know how many there are — or, far more often, it wants the values of some unknown constant that force there to be two, or one, or none. Solving the equation would be pointless work; you need a test that counts.
The test is already sitting inside the quadratic formula you derived in the "Solving a quadratic equation" section:
Every root the equation has comes from that expression, so everything about how many roots there are must be hiding in it. And there is only one place it can hide: the square root.
- If is positive, its square root is a real, non-zero number. Adding it and subtracting it give two different answers, so there are two distinct real roots.
- If is zero, then , and the does nothing at all: and are the same number. There is one root, and because the two roots have merged we call it a repeated root (or "equal roots", or "a double root" — all three phrases appear in past papers).
- If is negative, there is no real number whose square is negative, so the square root does not exist and there are no real roots at all.
That one expression therefore discriminates between the three possibilities, which is where its name comes from.
The discriminant
Read it off the equation only once it is arranged as ax² + bx + c = 0.
The same three cases, seen on the graph
Everything above was algebra. Here is the picture, which is often faster to reason with.
The roots of are the -values where the curve meets the -axis. A parabola can meet a horizontal line in three ways and no more: it can cut it twice, it can touch it once, or it can miss it entirely. Those are exactly the three discriminant cases, in the same order.
The middle case is worth dwelling on, because questions lean on it constantly. If the curve merely touches the axis, the axis is a tangent to the curve at that point, and the point of contact is the vertex. That is the geometric meaning of a repeated root.
The same parabola slid upwards. Only the constant changes, and with it the number of times the curve crosses the x-axis — which is exactly what the discriminant counts.
A clean demonstration
Work out the discriminant of three quadratics and say what each one tells you. Nothing here needs solving.
(i) . Read off , , :
, so two distinct real roots. (Bonus: is a perfect square, which means the surd disappears and the roots are rational — a reliable sign that the quadratic will factorise. It does: .)
(ii) . Here , , :
, so one repeated root. Indeed , so the only root is , and the curve touches the axis there.
(iii) . Here , , :
, so no real roots — the curve floats entirely above the -axis. You can confirm that independently by completing the square: , whose minimum value is positive.
Watch the two sign traps in (i) and (ii). In (i), was negative, so became and the discriminant grew. In (ii), was negative, but is positive — always bracket the substitution.
Where the repeated root actually sits
When the formula collapses to
so the repeated root is at — which is the axis of symmetry from the first section. That makes sense: the two roots have slid together until they met, and by symmetry they can only meet on the mirror line.
This is worth remembering because tangency questions usually have a second part: "…and find the coordinates of the point where the line touches the curve." You do not have to solve the quadratic again from scratch. Once you know the constant, either put it back in and factorise (the quadratic will be a perfect square) or go straight to , then substitute into the line to get .
Discriminant | Real roots | The graph |
|---|---|---|
two distinct | crosses the axis twice | |
one repeated (equal roots) | touches the axis | |
none | misses the axis entirely | |
at least one | crosses or touches |
“Real roots” includes the repeated case, so it means ⩾ 0, not > 0.
Tangent means equal roots
"The line is a tangent to the curve" is not a geometry statement you have to prove — it is a discriminant statement. Substitute, collect, and set . Similarly "touches" means and "does not meet" means .
There is a second route to tangency, using differentiation: set the gradient of the curve equal to the gradient of the line. Mark schemes accept it, and for some curves it is quicker. But the discriminant route needs no calculus and works on circles too, so make it your default.
Where it actually gets used
Almost nobody is asked for the discriminant of a quadratic that is handed to them. The real question is a line and a curve, with an unknown constant in one of them, and you are asked which values of the constant give two intersections, or one, or none.
The reason this works is the substitution idea. Where a line meets a curve, both equations hold at once, so setting them equal and collecting gives one quadratic whose roots are the -coordinates of the intersections. Counting intersections is therefore counting roots, and counting roots is what the discriminant does.
One curve, three lines of the same gradient. Setting the two equations equal produces a single quadratic whose discriminant decides which of the three pictures you are in.
The question says… | It means | So write |
|---|---|---|
“meet at two distinct points”, “intersects twice”, “cuts the curve twice” | two distinct roots | |
“is a tangent to”, “touches”, “meets at exactly one point”, “equal roots”, “a repeated root” | one repeated root | |
“do not meet”, “does not intersect”, “has no real roots”, “no solutions” | no real roots | |
“meets the curve”, “the roots are real”, “meet for all values of ” | at least one root |
Translate the words into a symbol before touching any algebra. Getting this line wrong scores zero with no recovery, however good the working after it.
- 1
Eliminate one variable. Usually substitute the line's expression for into the curve; if the curve has or but no , eliminating may be less work.
- 2
Collect into a three-term quadratic equal to zero, treating the unknown constant as an ordinary number. This line alone is usually worth a mark.
Everything after this depends on getting a, b and c right — write them down explicitly before going on.
- 3
Write , and down separately, brackets and all. If the unknown constant appears in more than one of them, that is normal.
For 2x² + (k − 3)x + 8 = 0 the value of b is the whole bracket k − 3, not k. Skipping this line is the commonest single error in the section.
- 4
Apply the condition that matches the wording: , or .
- 5
Solve the resulting equation or inequality in the constant — it is usually itself a quadratic, so find the critical values and then decide between "between" and "outside" as in the "Quadratic inequalities" section.
If the unknown is in the x² coefficient, the equation might not be quadratic
When a question hands you something like and asks for two distinct real roots, the discriminant condition is only half the story. If the term vanishes and the equation becomes the linear equation , which has exactly one root, not two.
So a fully careful answer excludes . In practice the excluded value often falls outside the discriminant answer anyway — but check, and say so if it does not. The same caution applies whenever a question writes the leading coefficient with a letter in it: , , have all appeared in the last five years.
The simplest possible discriminant question
Find the set of values of for which the equation has no real roots.
Show full working
- 1
The equation is already in the form , so read the coefficients straight off:
Write them down even when it feels unnecessary. It is the habit that saves you when b turns out to be a bracket.
- 2
Translate the wording. "No real roots" means :
- 3
Substitute the three coefficients:
- 4
Work out the numerical part: .
- 5
This is now an ordinary quadratic inequality in , solved exactly as in the "Quadratic inequalities" section. Its critical values come from , i.e. , so
Both square roots. A very large negative k also makes k² big, which is exactly why the answer is two-sided.
- 6
The curve is an upward parabola crossing at and , and we want it below zero — which is between the critical values:
- 7
Keep the inequalities strict, because the question said "no real roots" and at the discriminant is exactly zero, which is a repeated real root.
(equivalently )
The mark scheme is explicit that the first mark is for using — writing out the whole quadratic formula and stopping does not earn it. Show the discriminant on its own line, as a separate expression.
Line and curve: the standard four marks
A curve has equation and a straight line has equation , where is a constant.
Find the set of values of for which the curve and the line do not meet.
Show full working
- 1
Where the line meets the curve, the two -values are the same, so set the right-hand sides equal:
This single equation encodes 'the point is on both graphs'. Its roots are the x-coordinates of the intersections.
- 2
Collect everything on one side. Choose the side that makes the coefficient positive — here that means moving everything to the left, since becomes when it crosses over. Add , subtract , subtract :
Positive x² coefficient now means you will not have to flip an inequality later. It costs nothing to arrange it this way.
- 3
Collect the two terms by factorising out of them, and simplify the constants:
- 4
Write down the coefficients explicitly:
b is the whole bracket k − 3, not k. This is the single most penalised slip in the topic, and writing the line out is what prevents it.
- 5
"Do not meet" means no real roots, so the condition is :
- 6
Evaluate the numerical part: .
- 7
Add to both sides:
- 8
Undo the square. A square is less than exactly when the thing squared lies between and :
Two-sided, as in the inequalities section. Writing k − 3 < 8 alone gives k < 11 and silently drops the lower bound.
- 9
Add to all three parts of the chain:
- 10
Check the direction with a test value. Take , comfortably inside: the quadratic becomes , whose discriminant is — no intersections ✓.
One substitution, ten seconds, and it rules out the answer being the complement of what you wrote.
The published mark scheme awards nothing for using "", and refuses the final mark if you write signs or join the two ends with "or" — its guidance reads "CWO. Do not allow 'or'. A0 if sign or signs used." "Do not meet" is strict, and it is a single interval, so write it as one chain.
When the unknown sits in the x² coefficient
Find the set of values of the constant for which the quadratic equation
has two distinct real roots.
Show full working
- 1
Read off the coefficients. All three now involve or are affected by it:
Note a = 3k, not 3, and b is the whole bracket. Both are easy to misread when the letter is buried in the coefficient.
- 2
"Two distinct real roots" means :
- 3
Expand the square first, on its own:
- 4
Now the other part: . So the inequality is
- 5
Collect the terms: .
- 6
Find the critical values from . Two numbers multiplying to and adding to : both must be negative, and and work.
- 7
Upward parabola in , and we want it above zero, so the answer is outside the critical values:
Above zero ⇒ outside. Note this is a genuine 'or' answer — writing 4 < k < 16 would describe exactly the values that fail.
- 8
Finally, check the hidden condition. The question calls it a quadratic equation, so , i.e. , i.e. . Since , the value lies inside our answer set and strictly ought to be excluded.
At k = 0 the equation collapses to 8x + 3 = 0 — one root, not two. Say this even though the mark scheme's stated answer is the plain inequality; it costs one line and shows you understood.
- 9
Report the answer as the mark scheme does, noting the exclusion:
or
The published guidance says "B0 for use of and/or ." "Two distinct" roots is strict. And whenever the coefficient of contains the unknown, add "" to your mental checklist before you write the final line.
Tangency: find the constant, then find where it touches
A line with equation is a tangent to the curve with equation .
Find the possible values of the constant , and the corresponding coordinates of the points at which the line touches the curve.
Show full working
The two answers drawn to scale: both lines pass through (0, −6), and each touches the curve at exactly one point.
- 1
Set the two expressions for equal, since at a point of contact both equations hold:
- 2
Collect everything on the left. Subtract and add to both sides:
- 3
Factorise out of the two terms and add the constants:
−4x − mx = −(4 + m)x. Pulling the minus sign outside the bracket keeps the next step readable.
- 4
Coefficients:
- 5
"Is a tangent to" means the line touches the curve exactly once, which is a repeated root, so :
- 6
Squaring kills the outer minus sign, so , and :
- 7
Add :
- 8
Square-root both sides, taking both signs:
- 9
Deal with the two cases separately. If then . If then .
The question said 'possible values', plural. Losing the negative branch here loses half the marks.
- 10
Now the points of contact. Take first and put it back into the quadratic :
- 11
This must be a perfect square, because the discriminant is zero — and it is:
A perfect square is the signature of a repeated root. If your quadratic does not factorise as a square here, something earlier went wrong.
- 12
Find from the line, which is much less arithmetic than the curve: so the first point of contact is .
- 13
Now . The quadratic becomes
- 14
Again a perfect square:
- 15
And from the line : so the second point of contact is .
- 16
Shortcut check. For a repeated root, . With : ✓. With : ✓.
Under time pressure this is the fastest route to the contact point — no factorising needed at all.
touching at , and touching at
Keep each value of with its own point. The mark scheme awards the two points together and offers only a single special-case mark for one correct pair, so mixing up which goes with which costs more than it looks.
“Show that they always meet” — a discriminant that cannot be negative
Show that the curve with equation and the line with equation meet for all values of the constant .
Show full working
- 1
Rearrange the line to make one variable the subject. The curve has and but no , so making the subject is the cheaper move:
Always rearrange the linear equation, never the quadratic — the simultaneous-equations section explains why.
- 2
Substitute into the curve equation wherever appears:
- 3
Expand the bracket. Take care with the two minus signs: and .
- 4
Collect the terms:
- 5
Coefficients:
- 6
"Meet" means at least one real root, so the thing to examine is the discriminant:
- 7
Evaluate each piece. — square the as well as the . And , so subtracting it adds :
- 8
Now argue. Whatever real value takes, , so , so
This is the whole point of the question. You are not solving for k — you are showing that no k can ever make Δ negative.
- 9
Since for every value of , the quadratic always has two distinct real roots, so the line and the curve always meet — in fact they always meet twice.
Finish with the conclusion in words. A 'show that' is not complete until you have said what the algebra proves.
, and since we have for all , so the line and curve always meet.
"Show that … for all values of " always ends the same way: reach an expression like , or , and point out that a square is never negative. Look for that shape as soon as you see the phrase.
"Does not meet"
"Does not meet"
Mark schemes score this zero with no recovery. Translate the wording into a symbol before you touch any algebra.
or
The quadratic in k opens upwards, so “less than zero” is the region between the critical values, never outside them.
For , taking
, so
b is the entire coefficient of x after collecting, brackets and all.
Finding critical values and and stopping there
Stating the inequality
The critical values are typically one mark; the correctly-directed inequality is a separate one.
Writing only inside the quadratic formula
Write the discriminant out as its own expression
Mark schemes say 'Use of b² − 4ac but not just in the quadratic formula' and 'Not in quadratic formula unless b² − 4ac is isolated'. The method mark is for isolating it.
For , taking
The unknown lives in the leading coefficient. This also means the equation is only quadratic when k ≠ 0.
Finding from and stopping
, so or
Undoing a square always gives two branches. Tangency questions are written precisely so that both exist.
Finding the tangent constant but not the point of contact
Substitute back: the quadratic becomes a perfect square, or use
Tangency questions almost always ask for the point too, and it is worth two of the marks. Get y from the line, not the curve.
Discriminant questions are worth nearly twice as much as a typical completing-the-square part because they bundle four skills: eliminate, collect, apply the condition, solve an inequality. Most are phrased as a line meeting a curve or a circle, which is why this section is also a coordinate-geometry section. Tangency ("") is the most common condition, ahead of "" and "", and it usually comes with a second part asking for the point of contact.
Your turn
The first is a line meeting a curve; the second is the line-meets-circle version that coordinate geometry keeps asking for; the third is a 'show that they always meet'; the fourth is the 'true for all x' variant. Translate the wording into a symbol before you start each one.
- 1
A line has equation and a curve has equation , where is a constant.
Find the set of values of for which the line and the curve meet at two distinct points.
Stuck? Show hint
Set the two expressions for equal, collect to zero, and be careful: the coefficient of in the resulting quadratic is not .
Show solution
- 1
Set the two expressions equal:
- 2
Collect everything on the left. Subtract and subtract :
- 3
Factorise out of the middle terms and simplify the constants:
- 4
Coefficients: , , .
Using capital letters for the quadratic's coefficients avoids a clash with the b in the question — a small habit that prevents real confusion.
- 5
"Two distinct points" means two distinct real roots, so :
- 6
Evaluate the numerical part:
- 7
Add to both sides:
- 8
A square is greater than when the thing squared is further from zero than in either direction — so this splits into two cases, joined by or:
Greater-than gives 'outside', less-than gives 'between'. Getting these the wrong way round is the classic error.
- 9
Add to each:
- 10
Check with : the quadratic is , discriminant ✓ two roots. And with : , discriminant ✓ correctly excluded.
Answeror
- 1
- 29709/11 M/J 2022 Q9(b)6 marks
The equation of a circle is .
Find the set of values of the constant for which the line with equation intersects the circle at two distinct points.
Stuck? Show hint
A circle is not a function, but the method is unchanged: substitute the line's into the circle equation and collect into a quadratic in . Expect the coefficient to involve .
Show solution
- 1
Substitute into the circle equation everywhere appears:
- 2
Expand on its own first:
- 3
Expand . Now write everything out:
- 4
Collect the terms: .
- 5
Collect the terms: .
Three separate x terms, two of them carrying k. Gather them in one deliberate line rather than in your head.
- 6
Collect the constants: . So
- 7
Coefficients: , , . Note that can never be zero, since — so this really is a quadratic for every .
- 8
Two distinct intersections means :
- 9
Expand the square:
- 10
Expand the other part: . So
- 11
Collect. The constants cancel: . And .
- 12
There is no constant term, so factorise rather than divide. The common factor is :
Never divide an inequality by k — its sign is unknown, so you would not know whether to flip. Factorise and use critical values.
- 13
Critical values from :
- 14
The quadratic opens upwards, and we want it above zero, which is outside the critical values:
- 15
Write it as two pieces joined by or. The mark scheme's guidance is explicit: "Do not accept ."
That interval is precisely the set of k for which the line misses the circle — the exact opposite of what was asked.
Answeror
- 1
- 39709/12 F/M 2023 Q14 marks
A line has equation and a curve has equation , where is a constant.
Show that the line and the curve meet for all values of .
Stuck? Show hint
Form the intersection quadratic and find its discriminant in terms of . "Meet" allows touching, so you need for every — look for a perfect square.
Show solution
- 1
Set the two expressions for equal:
- 2
Collect everything on the left. Subtract and add :
- 3
Group the terms: .
This three-term quadratic is the first mark. Keep the constant term as the bracket 2 + 2k — it all belongs to c.
- 4
Coefficients:
- 5
Write the discriminant on its own line:
The mark scheme will not give the method mark if b² − 4ac only appears inside the quadratic formula.
- 6
Expand each part. , and :
- 7
Collect like terms: and .
- 8
Recognise a perfect square:
- 9
A square is never negative, so for every value of . So the discriminant is never negative, the quadratic always has at least one real root, and the line and the curve meet for all values of .
The last mark is for this conclusion in words. When k = 1 the discriminant is 0 and the line is a tangent — that still counts as meeting.
Answerfor all , so the line and curve always meet.
- 1
- 49709/13 M/J 2023 Q24 marks
The function is defined for by , where is a constant. It is given that for all values of .
Find the set of possible values of .
Stuck? Show hint
" for all " says the whole curve lies above the horizontal line . There are two ways to say that algebraically — one uses the minimum value, the other uses the discriminant of .
Show solution
- 1
Route 1 — the minimum value. The curve lies entirely above exactly when its lowest point is above . So find the minimum by completing the square.
- 2
Half of is , and , so
- 3
Add the back:
- 4
The square is never negative, so the minimum value of is , occurring at .
- 5
Require that minimum to be greater than :
- 6
Add to both sides:
- 7
Route 2 — the discriminant. " for all " means for all , i.e. the curve never touches or crosses the -axis. For an upward parabola that means no real roots:
Both routes are in the mark scheme. Route 2 is the one to reach for when the quadratic does not complete nicely.
- 8
Evaluate and :
- 9
Remove the bracket — the minus sign changes both signs inside:
- 10
Add to both sides: . Divide by (positive, so no flip): Same answer, as it must be.
- 11
Strict inequality: at the minimum is exactly , so at — and the question demanded for all .
The 'for all x' phrasing is what forces the strictness. Check the boundary case explicitly whenever you see it.
Answer - 1
Simultaneous equations — one linear, one quadratic
“
solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic, e.g. x + y + 1 = 0 and x² + y² = 25, 2x + 3y = 7 and 3x² = 4 + 4xy.
What the question is really asking
Two equations, two unknowns, and one of them is quadratic. To solve them simultaneously is to find every pair that satisfies both at once.
Geometrically, the linear equation draws a straight line and the quadratic one draws a curve, and a pair satisfying both is a point lying on both — an intersection. So there are usually two answers, sometimes one (the line is a tangent), sometimes none. And crucially, each answer is a pair of numbers, not a number. A question that says "find the coordinates" is not satisfied by a list of -values.
There is one method in the syllabus and it always works: substitution, always starting from the linear equation.
Why you rearrange the linear one
Both equations contain two unknowns, which is one too many to solve anything. The plan is to use one equation to express one unknown in terms of the other, then substitute that into the second equation, leaving a single equation in a single unknown.
The linear equation is the one to rearrange, for two reasons:
- It is easy. Making the subject of takes one line: . Making the subject of would require solving a quadratic — you would have made the problem harder, not easier.
- It cannot create false solutions. Rearranging a linear equation is reversible; every step can be undone. Squaring, or dividing by something that might be zero, is not.
Which variable to make the subject is worth two seconds of thought. Look at the quadratic: if it has but no , eliminate ; if it has and but no , eliminate . Choosing the one that avoids squaring a bracket saves real time.
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Rearrange the linear equation to make one variable the subject. Pick whichever is cheaper — from it is usually easier to take , but if the quadratic contains and , making the subject can be neater.
- 2
Substitute into the quadratic, so the whole equation is in one variable.
- 3
Expand and collect into a three-term quadratic .
This is the mark-earning line — mark schemes award it explicitly for “simplifying to a 3-term quadratic”.
- 4
Solve by factorising or the formula.
- 5
Substitute each solution back into the linear equation to get its partner, then present the answers as coordinate pairs.
Back into the LINEAR one: less algebra, and it cannot introduce spurious solutions the way the quadratic can.
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Check one pair in the original quadratic equation.
It is the only way to catch a sign error made during the substitution, and it takes fifteen seconds.
A clean demonstration
Solve simultaneously
Step 1 — the linear equation already has as the subject, so there is nothing to rearrange: .
Step 2 — substitute that into the quadratic. Everywhere the quadratic says , write :
Step 3 — collect into a three-term quadratic equal to zero. Move everything to the side that keeps positive, i.e. the right. Subtract and subtract from both sides:
Step 4 — solve. Two numbers multiplying to and adding to : they are and .
Step 5 — find each partner from the linear equation.
When : .
When : .
Step 6 — present as coordinate pairs, keeping each with its own :
Check the first pair in the quadratic: ✓, which matches .
The demonstration drawn to scale: the line y = x + 1 crosses the curve y = x² − 2x − 3 at the two solution pairs, (−1, 0) and (4, 5).
This is the engine of coordinate geometry
Substituting a line into a circle and collecting into a quadratic is exactly the same routine, and it is the core of many coordinate-geometry questions. Every line-and-curve discriminant question in "The discriminant" section also begins with this substitution.
Choosing which variable to eliminate
Find the coordinates of the points of intersection of the curve and the line with equations
Show full working
- 1
Look at the quadratic first: . It contains and but no — so only ever appears to the first power. Eliminating will therefore not create any new squares, and that is much less work.
Two seconds of looking before you start. Eliminating y here would mean substituting a bracket into 5y², which triples the algebra.
- 2
Rearrange the linear equation to make the subject. From , subtract and :
- 3
Divide by :
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Substitute into . The term becomes :
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The at the front cancels the in the denominator, which is why making the subject was worth it:
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Expand the bracket: and .
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Collect the terms: .
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Subtract from both sides to get everything on one side:
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Every term is divisible by , so divide through:
Dividing out a common factor before factorising makes the search almost instant — you are now looking for factors of 6, not 24.
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Factorise: two numbers multiplying to and adding to are and .
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So or .
- 12
Find each partner from the linear equation . With :
Back into the LINEAR equation, always. Substituting into the quadratic can hand you both partners for one y and you will not know which belongs.
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With :
- 14
Present as coordinate pairs, keeping each with the that produced it:
- 15
Check the second pair in the original quadratic: ✓
and
The question said coordinates, so a list of -values or -values is not an answer. When a question names its output format — "coordinates", "the set of values", "in the form " — that phrasing is what the final mark is for.
With unknown constants in the equations
The equation of a curve is and the equation of a line is , where and are constants.
Given that and , find the coordinates of the points of intersection of the curve and the line.
Show full working
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Put the given values in first, so you are working with numbers rather than letters. With the curve becomes and with the line becomes
Substitute the constants at the start, not the end. Carrying k and p through the algebra doubles the chances of a slip.
- 2
The line already gives in terms of , so substitute it into the curve wherever appears:
- 3
Expand the bracket. and :
- 4
Collect the terms: .
- 5
The coefficient is negative, so multiply every term by — this is an equation, not an inequality, so nothing needs turning round:
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Every term is divisible by :
Simplifying twice here — the sign and the common factor — turns an intimidating quadratic into a two-second factorisation.
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Factorise by splitting the middle term. and ; two numbers multiplying to and adding to are and :
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Factorise in pairs: from take ; from take :
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Pull out the common bracket:
- 10
So or , giving
- 11
Find each from the line . With :
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With :
- 13
So the points of intersection are
The mark scheme insists fractions are simplified. −11/2 + 3 must be finished to −5/2.
and
Notice the shape of the mark scheme: one mark for substituting and eliminating, one for reaching a three-term quadratic, and two for the coordinates. Even if the arithmetic falls apart, getting to that three-term quadratic banks half the marks — so write it down clearly on its own line.
One intersection: simultaneous equations meeting the discriminant
The straight line meets the curve at a single point .
(a) Find the value of the constant . [4]
(b) Find the coordinates of . [2]
Show full working
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(a) Substitute the line into the curve. The line gives , so replace :
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Expand the square on its own line:
- 3
Multiply that by :
- 4
So the equation is
- 5
Collect the terms and bring across:
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Coefficients: , , . Note that the unknown sits in the constant term.
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"Meets at a single point" means exactly one root, i.e. a repeated root, so :
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Evaluate the pieces: and .
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Expand the bracket carefully — the minus sign multiplies both terms: and .
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Simplify the constants: .
- 11
Add and divide by :
- 12
(b) Put back into the quadratic from part (a):
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Divide every term by :
- 14
Because the discriminant is zero this must be a perfect square, and it is:
One root, repeated — exactly what 'a single point' promised. If it had not factorised as a square, part (a) would be wrong.
- 15
Find from the line:
- 16
So is . Check in the curve: ✓
The check confirms both parts at once, which is why it is worth the twenty seconds on a six-mark question.
(a) (b) is
This is the standard two-part design: part (a) uses the discriminant to pin down the constant, part (b) puts it back and solves. Part (b) is nearly free once (a) is right — and the mark scheme awards only a single special-case mark for writing without an attempt at solving the quadratic.
Solving for and giving that as the answer
Giving the pairs and
Half a coordinate is not a point of intersection.
Substituting the -values back into the quadratic equation
Substituting back into the linear equation
The quadratic can hand you both partners for a value, so you end up with pairs that are not actually intersections.
Rearranging the quadratic equation and substituting into the linear one
Always rearrange the linear equation
Making x the subject of a quadratic requires solving it — you have made the problem harder, not easier.
Substituting into as
(x + 5)² is not x² + 25. Expand the square on its own line before multiplying by anything.
Leaving the answer as
Mark schemes state 'fractions must be simplified'. An unfinished arithmetic step is an unfinished answer.
Questions that ask only for a pair of simultaneous equations are not common, but they are worth more marks per part than anything else in the topic. The skill itself is everywhere: every "where does the line meet the circle", every "find the coordinates of the points of intersection", and every line-and-curve discriminant question opens with exactly this substitution. Most of these questions sit in coordinate geometry.
Your turn
Each pair needs a different choice about which variable to eliminate, and the last one links to the discriminant. Always finish with coordinate pairs, and check one pair in the original quadratic.
- 1
Solve simultaneously giving your answers as coordinates.
Show solution
- 1
Both equations already give , so set the right-hand sides equal:
- 2
Collect on the left, keeping positive. Subtract and add :
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Simplify: and .
- 4
Factorise: two numbers multiplying to and adding to are and .
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A repeated root, so there is only one intersection: . The line is a tangent to the curve.
A repeated root is not a mistake. It is the geometry telling you the line touches rather than crosses.
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Find from the line: .
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Check in the curve: ✓. The single point of intersection is .
AnswerOne point of intersection, — the line is a tangent to the curve.
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- 29709/12 O/N 2011 Q4(i)4 marks
The equation of a curve is and the equation of a line is , where is a constant.
(i) In the case where , find the coordinates of the points of intersection of the line and the curve.
Stuck? Show hint
The curve has but no , and appears only to the first power. Which variable is cheaper to eliminate?
Show solution
- 1
The curve contains but only to the first power, so eliminate — that avoids squaring anything.
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Rearrange the linear equation to make the subject. From , subtract :
- 3
Substitute into the curve equation :
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Expand the bracket: and .
- 5
Collect on the left with the constant moved across. Subtract :
16 − 13 = 3. Write the terms in descending powers before factorising — it makes the pattern visible.
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Factorise: two numbers multiplying to and adding to are and .
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So or .
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Find each from the linear equation . With : . With : .
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So the intersections are and .
- 10
Check in the curve: ✓
The mark scheme wants all four coordinates. A check on one pair costs almost nothing and confirms the pairing.
Answerand
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- 39709/13 O/N 2022 Q10(a)4 marks
The diagram shows the circle and the straight line intersecting at the points and . The point on the -axis is such that is perpendicular to the -axis.
(a) Find the coordinates of .
Fig. 10.1
Stuck? Show hint
The line already gives , so substitute it into the circle. You will get two values of — use the diagram to decide which one belongs to .
Show solution
- 1
Substitute into :
- 2
Expand the square on its own:
- 3
So
- 4
Collect everything on the left: and .
Reaching this three-term quadratic earns the first two marks.
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Factorise by splitting the middle term. and ; two numbers multiplying to and adding to are and :
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In pairs: from take ; from take :
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So or .
- 8
In the diagram, is the intersection with positive (above , on the positive -axis), so has .
The other root, x = −1/5, belongs to B. Using the diagram to choose the root is part of the question.
- 9
Find from the line: . Check in the circle: ✓
Answeris
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- 4
A line has equation and a curve has equation .
Find the set of values of for which the line and the curve do not meet.
Stuck? Show hint
"Do not meet" is a discriminant condition, but you still have to do the substitution first. Be careful: the line's gradient ends up inside .
Show solution
- 1
Set the two expressions for equal:
- 2
Collect everything on the left, keeping positive. Subtract and add :
- 3
Simplify the constants:
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Write down the coefficients: , , .
b is −m, not m. It will not matter here because b gets squared, but forming the habit is what protects you when it does matter.
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"Do not meet" means no real roots, so :
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, and :
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Critical values from : and .
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Upward parabola in , and we want it below zero, so the answer is between the critical values:
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Check with , which is inside: the line is horizontal, and the curve has minimum value , so they never meet ✓.
Picking a test value you can reason about geometrically is the strongest possible check.
Answer - 1
Equations that are quadratic in something else
“
recognise and solve equations in x which are quadratic in some function of x, e.g. x⁴ − 5x² + 4 = 0, 6x + √x − 1 = 0, tan²x = 1 + tan x.
The pattern to look for
Look at these three:
None is a quadratic in . All three are quadratics in something: in , in , and in .
The pattern is always the same shape:
and the test for whether you are looking at it is one question: is one of the powers exactly the square of another?
- is the square of ✓
- is the square of ✓
- is the square of ✓
- is the square of ✓
- is the square of ✓
- is the square of ✓
Once you see it, the question becomes a routine you already know.
Why the substitution is legitimate
Write for the inner thing. Then becomes
which is a perfectly ordinary quadratic. There is nothing magic here: you have simply given a name to a repeated expression. The equation has not changed, only the way it is written.
But the substitution is only half the job, and the second half is where the marks are lost. The question asked for , not for . Solving the quadratic gives values of ; each of those is then a new equation to solve, and the number of -values it produces depends entirely on what stood for:
- If , then gives two values of (namely ) when , one when , and none at all when .
- If , then gives exactly one real value of for every , positive or negative, because cube roots of negatives exist.
- If , then cannot be negative, so any negative root of the quadratic must be thrown away; each surviving gives exactly one .
- If or , then must lie between and , and each surviving value typically gives several angles in the interval.
- If , any real is allowed, and each gives one angle per of interval.
So the return journey needs as much care as the outward one. Mark schemes reflect this: on one recent question the final mark is annotated WWW ("without wrong working"), which means leaving and as your answer scores nothing for it.
A clean demonstration
Solve .
Step 1 — spot it. The powers are and , and . So this is a quadratic in .
Step 2 — substitute. Let . Then , and the equation becomes
Step 3 — solve the quadratic in . Two numbers multiplying to and adding to : both negative, and and work.
Step 4 — go back. These are not the answers; the question asked for . Replace by , giving two separate equations:
Step 5 — solve each, remembering both square roots each time.
Four solutions: .
Check one: gives ✓
The point of this demonstration is step 5. A quadratic has at most two roots, but a quartic can have four, and losing the negatives is by a distance the most common error in this sub-topic.
The substitution is the easy half. The step people lose marks on is the return journey: u = 1 and u = 4 are not solutions of the original equation, and each one can produce more than one x.
Equation | Substitute | Becomes | Going back |
|---|---|---|---|
gives two for each positive ; none if | |||
one real cube root per , negatives included | |||
(after ) | , so reject negative ; then | ||
(after ) | one real cube root per | ||
is a square, so reject negative ; then | |||
any real allowed; one angle per | |||
need ; then all angles in the interval | |||
; then , then the angles |
Disguises that have appeared on recent Paper 1 questions. Note how different the last column is in each row — that is the part questions are really testing.
- 1
Spot the pattern. Is one power exactly the square of another? and ; and ; and ; and ; and .
- 2
Clear fractions first if you need to. An equation with becomes a quadratic in only after multiplying every term by .
Check first that the thing you are multiplying by cannot be zero. It never can here — the original expression would be undefined — so nothing is lost.
- 3
For a trigonometric equation, get everything into one function first. Replace by , or by , until only one trigonometric ratio remains.
You cannot substitute a single u while both sin θ and cos θ are present. This is usually the 'show that' part (a) of the question. One shape worth knowing: an equation of the form a sin θ = b cos θ divides straight through by cos θ to give tan θ = b/a — first check that cos θ = 0 is not itself a solution.
- 4
Substitute for the inner function and solve the quadratic in . You may work in or directly — a separate letter just makes the structure obvious, and is safer under pressure.
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Reject the impossible values of . A square cannot be negative; cannot be negative; and cannot leave .
Questions are built so that one root is impossible. Saying explicitly why you rejected it is often worth a mark on its own.
- 6
Go back. Replace and solve for — a separate equation for every surviving root.
This is the step that gets skipped, and it is where the final marks live.
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Count. Have you got every solution the interval allows, and no extras outside it?
Getting all the angles, and only those angles
Trigonometric equations are taught fully in the Trigonometry note; this is the minimum you need for the examples below. Trigonometric disguises finish with an equation like over a stated interval, and the mark scheme insists on all the solutions in that interval and no others. The routine:
- Take the inverse function on your calculator. That gives one angle, the principal value — in for , in for and .
- Generate the partner angles from the symmetry of the graph:
- : if is a solution, so is , then add or subtract .
- : if is a solution, so is , then add or subtract .
- : if is a solution, so is , and so on every .
- Keep only those inside the interval the question gave — and check both ends of it.
Two habits: work in degrees unless the interval is written with in it, and never round an intermediate angle before generating its partners.
Disguise ① — a power, hidden behind a fraction
Solve the equation
Show full working
- 1
As it stands this is not a quadratic in anything, because of the fraction. Clear it: multiply every term by .
Before multiplying, check x³ cannot be zero. If x were 0 the original expression would divide by zero, so x = 0 was never a candidate and nothing is lost.
- 2
Term by term: ; ; and .
- 3
Now look at the powers: and , and . So this is a quadratic — in .
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Substitute , so that :
- 5
Factorise: two numbers multiplying to and adding to . Both must be negative, and and work.
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So or .
- 7
Go back, because the question asked for . Each value of gives its own equation:
- 8
Take the cube root of each. Unlike a square root, a real cube root is unique — there is only one real number that cubes to , and only one that cubes to :
No ± here. That is the difference between an even power and an odd one, and it is exactly what the disguise is testing.
- 9
Check in the original: ✓
and
Three marks, three moves: recognise, solve, return. The mark scheme awards them in exactly that order, and the third is marked WWW — "without wrong working" — so a stray left as the final answer loses it.
Disguise ② — a quartic, with one root that has to be thrown away
(a) Express in the form , where and are constants. [2]
(b) Hence find the exact solutions of the equation . [3]
Show full working
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(a) Take the out of the first two terms:
Part (a) on its own is an exercise in the "Completing the square" section — here the same moves feed part (b), which is where the disguise lives.
- 2
Complete the square inside. Half of is , and , so :
- 3
Multiply the through both terms: So and .
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(b) The word hence means part (a) is meant to be reused. Compare the two expressions: They are identical if , because then .
Spotting the match is the whole idea. Notice that x⁴ = (x²)², which is why x² is the right thing to call y.
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So substituting into part (a)'s answer, and the equation becomes
- 6
Add to both sides:
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Divide both sides by :
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Take the square root of both sides, with :
The ± here is what produces two separate cases. Dropping it loses the whole of the rest of the question.
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Deal with the two cases separately. If then
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If then
- 11
Reject . No real number squares to give a negative, so this case produces no solutions at all.
Say so explicitly. The mark scheme allows the −1 to be omitted only if the ±3 was clearly shown — the safest route is to write it down and dismiss it.
- 12
From , take both square roots:
- 13
The question said exact, so leave the surd. has no square factors and cannot be simplified.
- 14
Check : ✓
(a) (b)
Two lessons. First, when a part (a) is in a different letter from part (b), the letter is a hint: the examiner has already told you what substitution to make. Second, the mark scheme here adds "use of calculator with no working scores 0/3" — a bare answer earns nothing on a "hence" question.
Disguise ③ — a square root, then the same equation in tan x
(a) Solve the equation . [4]
(b) Hence solve the equation for . [3]
Show full working
- 1
(a) The repeated object here is — it appears once on the top and once on the bottom. Let and note straight away that , since a square root is never negative.
Record the restriction on u the moment you make the substitution. It is what tells you later which roots to keep.
- 2
In terms of the equation is
- 3
Clear the fraction by multiplying every term by . (We know , since would be undefined.)
- 4
Write it in the usual order:
- 5
Factorise by splitting the middle term. , and ; two numbers multiplying to and adding to are and .
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Factorise in pairs. From take ; from take :
- 7
Pull out the common bracket:
- 8
So or . Both are positive, so both survive the restriction — nothing to reject this time.
- 9
Go back. , so square each value to recover :
Squaring is the inverse of square-rooting here, and it is safe precisely because both u values were positive.
- 10
(b) The second equation is the first one with replaced by throughout. So from part (a),
Do not restart. 'Hence' means the algebra is already done; only the trigonometry is left.
- 11
Both values are positive, which is consistent with the original equation needing for the square root to exist. Positive tangent means is in the first or third quadrant.
- 12
Take . The calculator gives the principal value
- 13
Tangent repeats every , so the next solution is . Adding another would give , which is outside the interval. So from this value: and .
- 14
Now . The principal value is
- 15
Add : . So and .
- 16
Collect all four, each to 1 decimal place:
Four values, because two tangent values each give two angles in a 360° interval. Check you have not left any out and have added none outside.
(a) and (b)
This is the classic two-part design: an abstract equation in (a), the same equation wearing trigonometric clothes in (b). The whole of part (b) is worth three marks and takes ninety seconds if you reuse part (a). Restarting from scratch is allowed but will cost you the time you needed for question 10.
Disguise ④ — substituting for a whole bracket
By using a suitable substitution, solve the equation
Show full working
- 1
The repeated object is the whole expression : it appears once on its own and once in a denominator. So take
Substitute for the biggest repeated block you can see. Choosing u = 2x − 3 also works but gives a quartic in u instead of a quadratic — more work for the same answer.
- 2
Note the restriction: is something squared, so . In fact too, since it sits in a denominator.
- 3
In terms of :
- 4
Multiply every term by to clear the fraction:
- 5
Write it in descending powers:
- 6
Factorise: two numbers multiplying to and adding to are and .
- 7
So or .
- 8
Reject , because is a square and a square cannot be negative.
The question was built around this rejection. Stating the reason in words is what earns the mark, not just quietly dropping the value.
- 9
Go back with :
- 10
Take the square root of both sides, with :
- 11
Case : add to get , so
- 12
Case : add to get , so
- 13
Check : then , so , and the equation reads ✓
and
"By using a suitable substitution" is an instruction: write down what your is. And when you choose it, pick the largest repeated block — here rather than — because that is what turns the equation into a quadratic in one step rather than two.
Disguise ⑤ — a trigonometric fraction, and a root that is impossible
Solve the equation for .
Show full working
- 1
Everything here is in , so that is the object to treat as the variable. First clear the fraction: multiply every term by .
cos θ cannot be zero here, since 3/cos θ appears in the equation — so multiplying through is safe.
- 2
Term by term: ; ; .
- 3
Collect everything on the left so the equation equals zero:
- 4
This is now a quadratic in . If it helps, write to get — but note that must lie between and , because that is the entire range of the cosine function.
Recording −1 ⩽ u ⩽ 1 now is what makes the rejection obvious in two steps' time.
- 5
Factorise. and ; two numbers multiplying to and adding to are and :
- 6
In pairs: from take ; from take :
- 7
- 8
Reject . The cosine of a real angle never exceeds , so this gives no solutions.
Write the reason down. 'Reject cos θ = 3 since cos θ ⩽ 1' is exactly what the examiner wants to see.
- 9
Solve . The calculator gives the principal value
- 10
Cosine is an even function — — so is also a solution. Equivalently, the other solution in a full turn is , which is the same angle as .
- 11
Check both against the interval . Both and lie strictly inside it, so both are kept:
A negative interval is a deliberate trap. If you only look for angles between 0° and 360° you will find 60° and 300°, and report 300° — which is outside the interval.
and
Two things to carry forward. Trigonometric quadratics almost always have one impossible root — , , — and rejecting it with a stated reason is worth a mark. And always read the interval before you generate angles: the interval, not the quadratic, decides how many answers there are.
Disguise ⑥ — quartic in sine, four angles out
Solve the equation for .
Show full working
- 1
The two powers are and , and . So this is a quadratic in — note, in , not in .
There will therefore be an extra square root to undo on the way back. Getting this right is what produces four angles instead of two.
- 2
Let , remembering that (it is a square) and in fact :
- 3
Factorise. and ; two numbers multiplying to and adding to are and :
- 4
In pairs: from take ; from take :
- 5
- 6
Reject , since is a square and cannot be negative.
- 7
Go back one level:
- 8
Take the square root of both sides — both signs:
This is the step that doubles the number of answers. Keeping only the positive root halves your marks.
- 9
Solve first. The principal value is
- 10
Sine is positive in the first and second quadrants, and the second-quadrant partner is . Both lie in .
- 11
Now . Sine is negative in the third and fourth quadrants, and the related acute angle is still , so
- 12
All four lie inside the interval, and there are no others:
The mark scheme gives one mark for any two correct and the second only for all four with no extras — so a missing angle and a spurious one cost the same.
Count the layers before you start. is three steps back, and each of the middle two can double the number of answers. A quadratic in over a full typically produces four angles; a quadratic in produces two.
Disguise ⑦ — the “hence”, in full
(a) Use completing the square to find the exact solutions of the equation . [2]
(b) Hence solve the equation for . [3]
Show full working
- 1
(a) was worked in full in the "Solving a quadratic equation" section. In brief: take the out of the first two terms to get , complete the square inside to get , multiply out to , and solve:
- 2
(b) The word hence says part (a) is meant to be reused, so the job is to make part (b)'s equation look like part (a)'s. Start by clearing the fraction: multiply every term by .
tan θ cannot be zero, since 1/tan θ appears in the equation. So multiplying through loses nothing.
- 3
Term by term: ; ; .
- 4
Collect everything on the left:
- 5
Compare with part (a)'s equation . They are identical with replaced by — which is exactly what hence was pointing at.
Say this out loud. Once you see the match you can copy part (a)'s answers straight across instead of solving anything.
- 6
So
- 7
Turn each into a decimal, keeping plenty of digits. , so
- 8
First value. , which lies inside . So (1 d.p.). The next tangent solution would be , which is outside the interval.
- 9
Second value. . That is not in the interval, so it is not an answer as it stands.
The calculator always returns the principal value, which for tan⁻¹ lies between −90° and 90°. It is a starting point, not an answer.
- 10
Tangent repeats every , so add : which does lie inside the interval. So (1 d.p.).
- 11
Check there are no more. Adding another gives , outside; subtracting gives , outside. So exactly two solutions.
A quadratic in tan θ over a 180° interval gives exactly one angle per root. Two roots, two angles — a useful sanity check.
(a) (b) and
A "hence" part is a gift: it tells you the algebra is already done and only the substitution and the interval are left. Answering (b) from scratch is not penalised, but it costs several minutes you do not have. And note that part (a) demanded exact answers while part (b) wants decimals — read each instruction separately.
and — four solutions
Every positive value of x² gives two values of x. Losing the negatives is the most common error in this sub-topic.
, so and
is impossible, so only
Squaring both sides of a rejected value silently turns an impossible solution into a plausible-looking one.
Giving and as the final answer
and
The question asked for x. Marks for the substitution are conditional on returning from it — and the mark is often flagged WWW.
Reporting because the calculator said so
, adjusted into the given interval
The number of angles is set by the interval, not by the number of roots of the quadratic.
Dropping silently
Write “reject since ”
The rejection is usually worth a mark, and only if the reason is visible. Deleting the root without comment reads as an omission.
, giving
Undoing the square on sin²θ produces two equations, not one. Half the angles disappear if you take only the positive root.
Multiplying through by without comment when could be
Note that appears, so already
Multiplying by something that might be zero can add false solutions. Here the original equation rules it out, but you should know why you are allowed to.
This sub-topic is examined far more often than its one line in the syllabus suggests, and the shape of the question is stable: part (a) hands you a quadratic to solve or a form to complete, part (b) says hence and dresses the same quadratic as a trigonometric, surd or power equation. The marks are mostly for the return journey — going back to the original variable, rejecting an impossible value, finding every angle in the interval — not for the substitution. If part (a) of a question looks suspiciously easy, part (b) is about to reuse it.
Your turn
Two square-root disguises, a power, a trigonometric one and a bracket. In each case say out loud what your u stands for, and check at the end that you have solved for the variable the question actually asked about.
- 1
Solve the equation .
Stuck? Show hint
is the square of . Let — and remember that cannot be negative.
Show solution
- 1
Since , this is a quadratic in . Let , so and :
Record u ⩾ 0 straight away: the square-root sign always means the non-negative root.
- 2
Factorise by splitting the middle term. and ; two numbers multiplying to and adding to are and :
- 3
In pairs: from take ; from take :
- 4
So or .
- 5
Reject : cannot be negative.
If you squared −1/2 anyway you would get x = 1/4, and 6(1/4) + 1/2 − 1 = 1, not 0. Squaring hides the mistake, which is why the rejection must come first.
- 6
Go back with : , so square both sides:
- 7
Check: ✓
Answeronly
- 1
- 29709/12 O/N 2023 Q9(a)4 marks
The diagram shows curves with equations and . The curves intersect at points and .
(a) Find the coordinates of and .
Fig. 2
Stuck? Show hint
Set the two expressions equal, then multiply every term by . What you get is a quadratic in .
Show solution
- 1
At an intersection the -values are equal:
- 2
Multiply every term by . Since and :
means , so multiplying by clears it. cannot be here, because would be undefined.
- 3
Collect everything on the left:
- 4
Divide every term by :
- 5
This is a quadratic in . Let , so and :
- 6
Factorise: two numbers multiplying to and adding to are and .
- 7
Both are positive, so both are allowed. Go back: , so
- 8
Find each from the simpler curve, . At : .
- 9
At : .
- 10
Check in the other curve: ✓
Answers without working score 0 on this question, so the substitution and the quadratic must be shown.
Answerand
- 1
- 39709/12 M/J 2023 Q43 marks
Solve the equation .
Stuck? Show hint
is the square of . And remember that cube roots behave differently from square roots when the number is negative.
Show solution
- 1
The powers are and , and , so this is a quadratic in . Let :
- 2
Factorise by splitting the middle term. , and . Two numbers multiplying to and adding to : since the sum is nearly as large as the product, one number is small — try and . Product ✓, sum ✓.
When b is huge compared with a and c, look for a pair like 1 and (ac). Guessing sensibly beats listing every factor pair.
- 3
Split:
- 4
In pairs: from take ; from take :
- 5
Pull out the common bracket:
- 6
So or .
- 7
Go back to . First : the cube root of a negative number is negative, and , so
Do not reject the negative value. A cube root of a negative number is a perfectly good real number — this is where the odd power differs from an even one.
- 8
Then : since ,
- 9
Check : ✓
Answerand
- 1
- 49709/12 O/N 2022 Q3(b)3 marks
Solve the equation for .
Stuck? Show hint
Divide by the common factor first. Then remember that the interval only runs to , so each value of gives at most one angle — but check both ends of the interval carefully.
Show solution
- 1
Every coefficient is even, so divide through by before doing anything else:
Smaller numbers, easier factorisation, same equation. Always look for a common factor first.
- 2
This is a quadratic in . Writing , with :
- 3
Factorise: , ; two numbers multiplying to and adding to are and .
- 4
In pairs: from take ; from take :
- 5
- 6
Both values lie between and , so neither is rejected this time.
Check the range even when nothing fails. It takes a second and it is the habit that catches cos θ = 3 when it does appear.
- 7
Solve . The only angle in with cosine is and the interval includes , so it counts.
The interval is written with ⩽ at both ends, so the endpoints are allowed. An interval written with < would exclude 0°.
- 8
Solve :
- 9
Over to the cosine function decreases steadily from to , so each value of gives exactly one angle. There are no others.
Answerand
- 1
- 59709/11 O/N 2023 Q8(a)4 marks
The diagram shows the curves with equations and meeting at points and .
(a) By using the substitution find, by calculation, the coordinates of and .
Fig. 8.1
Stuck? Show hint
Set the two expressions for equal first. After the substitution you should have a quartic in that is a quadratic in — so there are two layers to come back through.
Show solution
- 1
Where the curves meet, the -values are equal:
- 2
Apply the given substitution . Then and :
- 3
Collect everything on one side:
- 4
This is a quadratic in , since . Factorise it treating as the variable: two numbers multiplying to and adding to are and .
- 5
Split the middle term into :
- 6
Factorise in pairs. From take ; from take :
- 7
Pull out the common bracket:
The mark scheme wants the factors (or formula, or completed square) shown — a bare answer from a calculator loses the method mark.
- 8
So or , i.e. .
- 9
Reject : a square cannot be negative.
- 10
From , take both square roots:
Both signs. This is the layer that produces the two points A and B — with only u = 1 you would find one point and lose two marks.
- 11
Now undo the substitution. , so
- 12
Finally find the -coordinates. Use the simpler curve, . Since in both cases, for both points.
Both points have the same y because y depends only on u², and u² is 1 either way. Noticing that saves a second substitution.
- 13
So the two intersection points are and .
Answerand are and
- 1
Where quadratics hide in the rest of Paper 1
The topic that is never the whole question
Everything so far has been the mechanics. This short section is about recognition — spotting a quadratic when the question is about something else, which is where many of the topic's marks actually are. The table lists the places it turns up most often, roughly from most to least common. You will meet each of these topics in a later note.
Where it turns up | What the quadratic looks like |
|---|---|
A line meeting a curve | the intersection quadratic; its discriminant decides the number of crossings |
Range and composition of functions | complete the square to get a minimum; a composite that expands into a quadratic |
Lines and circles | substitute a line into a circle; the discriminant tests tangency |
Straight lines | perpendicular gradients, or a distance, giving a quadratic in the unknown coordinate |
Trigonometric equations | a quadratic in , or , then all angles in the interval |
Arithmetic and geometric progressions | two conditions on an AP or GP, eliminating one unknown to leave a quadratic in , or |
Inverse functions | invert a quadratic by completing the square, then choose the right square-root branch |
Binomial expansion | a condition on the coefficients giving a quadratic in an unknown constant |
Gradients and stationary points | the gradient of a cubic is a quadratic; set it or |
Places in Paper 1 where a quadratic appears inside a question on another topic.
Whenever a Paper 1 question leaves you with one equation in one unknown, and that unknown appears squared, stop and treat it as a quadratic:
- Expand every bracket and clear every fraction.
- Collect all terms on one side, with a positive coefficient of the squared term.
- Factorise, or use the formula, or complete the square.
- Check each root against the context, and reject any that the situation forbids.
Step 4 is the one that separates the marks. In a progression, must be a positive whole number. In geometry, a length must be positive. In a convergent GP, . In a domain-restricted function, the root must lie in the domain.
A quadratic hiding inside an arithmetic progression
An arithmetic progression has first term and common difference . The th term is and the sum of the first terms is .
Find the values of and .
Show full working
- 1
Write down the two facts as equations. The th term of an arithmetic progression is , and here and the th term is :
Two facts, two unknowns. Getting both onto paper before doing any algebra is half the battle.
- 2
The sum of the first terms is . With and :
- 3
Simplify equation to get on its own. Expand: , so
- 4
Substitute that into . First tidy the bracket in : . Putting in gives
Simplify the bracket completely before multiplying by the 3N/2 outside. Doing both at once is where the arithmetic collapses.
- 5
So equation becomes
- 6
Multiply both sides by to clear the fraction:
- 7
Divide both sides by :
- 8
Expand the left-hand side:
- 9
Divide through by :
- 10
Collect to zero:
This three-term quadratic is worth its own mark in the mark scheme, before any solving.
- 11
Factorise. Two numbers multiplying to and adding to . One is negative and the positive one is bigger; and , so take and .
- 12
So or .
- 13
Reject . counts terms of a progression, so it must be a positive whole number. Hence
This is step 4 of the recognition rule. A quadratic in a real-world context nearly always has one impossible root, and stating why you reject it is part of the answer.
- 14
Finally, get from :
- 15
Check: the th term is ✓, and the sum of the first terms is ✓
and
This question is filed under Series, and its mark scheme never uses the word "quadratic" until the fourth line. That is the pattern: two conditions on a progression, eliminate one unknown, and a quadratic falls out. The same happens with geometric progressions, where the quadratic is usually in and the rejection is " for convergence".
Three signals that a quadratic is coming
- "Find the possible values of…" — plural, so the equation ahead has two roots. Almost always a quadratic.
- "Find the value of the constant " where appears in two different conditions — eliminate between the two conditions and a quadratic in appears.
- Any phrase about intersections, tangency or "does not meet" — you are heading for a discriminant, which means you are heading for a quadratic first.
And one signal that the answer needs checking: any question set in a physical or counting context. Lengths, radii, numbers of terms and common ratios all carry restrictions that one of your two roots will usually violate.
Your turn
Each of these belongs to another topic. Your job is to notice the quadratic, form it, solve it, and reject a root only when the question gives a reason.
- 1
The point has coordinates and the point has coordinates , where is a constant.
Given that the distance is units, find the possible values of .
Stuck? Show hint
Use the distance formula, and square both sides straight away so the surd disappears. "Possible values" is plural — expect two answers.
Show solution
- 1
The distance between two points is , so
- 2
Square both sides immediately, to get rid of the root:
Squaring first is always the right move with a distance condition. Working with a surd on the left just adds a step you would have to undo.
- 3
Evaluate the first bracket: and .
- 4
Subtract from both sides:
- 5
Take the square root of both sides, with :
Both signs. R could be well above P or well below it, and the question's word 'values' is telling you both exist.
- 6
Case : add to get .
- 7
Case : add to get .
- 8
Neither is ruled out — is just a -coordinate, so it may be negative. Both answers stand.
Step 4 of the recognition rule: check the context, and reject only if the context genuinely forbids a value. Here it does not.
- 9
Check : , so ✓
Answeror
- 1
- 2
The first term of an arithmetic progression is and the common difference is . The sum of the first terms is .
Find the value of .
Stuck? Show hint
Use , clear the fraction, and remember what has to be at the end.
Show solution
- 1
Substitute and into the sum formula:
- 2
Simplify inside the bracket: , and , so the bracket is .
- 3
Set that equal to :
- 4
Multiply both sides by to clear the fraction:
- 5
Expand the left-hand side:
- 6
Divide every term by :
Dividing out the common factor before collecting keeps the numbers small enough to factorise by inspection.
- 7
Collect to zero:
- 8
Factorise: two numbers multiplying to and adding to . Factor pairs of are , , , , . The pair and differ by , so take and .
- 9
So or .
- 10
Reject , because counts terms and must be a positive whole number. Hence
Step 4 of the recognition rule. Leaving −9 in the answer would cost the final mark even with all the algebra correct.
- 11
Check: ✓
Answer - 1
- 39709/13 M/J 2023 Q8(a)5 marks
A progression has first term and second term , where is a positive constant.
For the case where the progression is geometric and the sum to infinity is , find the value of .
Stuck? Show hint
In a geometric progression the common ratio is (second term) ÷ (first term), and the sum to infinity is . Expect a quadratic in — and read the question again before you give two answers.
Show solution
- 1
In a geometric progression each term is the previous one times the common ratio , so = second term ÷ first term:
Geometric progressions are taught in the Series note. Only two facts are needed here: r = second term ÷ first term, and the sum to infinity formula.
- 2
The sum to infinity of a geometric progression is . Set it equal to :
- 3
Simplify the denominator by writing as :
Combining 1 − a/(a + 2) into a single fraction first stops the compound fraction getting out of hand.
- 4
Dividing by is the same as multiplying by :
- 5
Multiply both sides by :
- 6
Expand and collect to zero:
This three-term quadratic is a method mark in the mark scheme.
- 7
Factorise: two numbers multiplying to and adding to . Since and , take and :
- 8
So or .
- 9
Reject , because the question says is a positive constant. Hence
The mark scheme accepts 22 only. Leaving −24 in the answer loses the final mark.
- 10
Check: , and ✓
Answer - 1
Everything on one page
General form — c is the y-intercept
The perfect square — the coefficient of x is TWICE the number in the bracket
Completed square form
Turning point, read straight off
Range from the completed square — a statement about y, never about x
Axis of symmetry — midway between the roots, and where a repeated root sits
The zero-product rule — why factorising solves anything
The quadratic formula
The discriminant
Two distinct / one repeated (tangent) / no real roots
Below zero (between) vs above zero (outside), for a > 0
Undoing a square inside an inequality — two-sided, always
The disguises — substitute, solve, then come all the way back
Can you do all of these?
Sketch from nothing: direction, -intercept, roots, vertex, range
Complete the square for without writing down the general formula
Complete the square for into the form , getting both signs right
State the vertex, minimum value and range of from the form alone — and write the range in terms of , not
Show that for all by completing the square, and say why in words
Factorise by splitting the middle term
Derive the quadratic formula by completing the square on
Solve without dividing by , and get both roots
Solve and write the answer with the right notation
Explain why has its solution between the roots
Say what must be for "is a tangent to", "does not meet" and "has real roots"
Turn "the line does not meet the curve" into an inequality in and solve it
Given that a line is a tangent to a curve, find the constant and the coordinates of the point of contact
Solve a linear-and-quadratic pair and give the answers as coordinate pairs
Solve and get all four roots
Solve and reject the impossible value of
Solve and explain why there is no at the end
Solve for and get all four angles
Solve for and check both angles are in range
State why must be rejected, in a sentence an examiner would accept