Quadratics Quick Revision — 9709 Paper 1

01

The shape and the vocabulary

Definitions

A quadratic in $x$ is an expression whose highest power of $x$ is $2$ :

ax^2 + bx + c, \qquad a \neq 0

$a$ is the leading coefficient (coefficient of $x^2$ ), $b$ the coefficient of $x$ , $c$ the constant term.
$a \neq 0$ is what makes it a quadratic. If a question puts an unknown in the leading coefficient — $3kx^2 + (k+8)x + 3 = 0$ — it is quietly requiring $k \neq 0$ .
Set it to zero and it is a quadratic equation. A value of $x$ satisfying it is a root = solution = zero; all three words appear in past papers.
Write $y = ax^2 + bx + c$ and it is a curve — a parabola: one smooth U, no corners, exactly one turning point, symmetric about a vertical line through it. That turning point is the vertex (a minimum if the curve opens up, a maximum if it opens down).
The roots are where the curve crosses the $x$ -axis, because crossing the axis is what $y = 0$ means. Every algebraic question in this topic has that picture behind it.

Three things you can read off without working

$c$ is the $y$ -intercept — put $x = 0$ and everything else vanishes.
The sign of $a$ is the direction: $a > 0$ opens upwards (minimum), $a < 0$ opens downwards (maximum).
The axis of symmetry is $x = -\dfrac{b}{2a}$ , and the vertex sits on it — equivalently, the vertex is exactly halfway between the roots.

Everything a Paper 1 question can ask you to find about a quadratic, marked on one curve. The curve is symmetric about the dashed line, so the vertex sits exactly midway between the two roots.

Form	Looks like	Hands you	Reach for it when
General	$ax^2 + bx + c$	the $y$ -intercept $c$	starting out, or you need the discriminant
Completed square	$a(x+p)^2 + q$	vertex $(-p,\ q)$ , and the range	maximum / minimum, range, sketching, “no real roots”
Factorised	$a(x-\alpha)(x-\beta)$	the roots $\alpha,\ \beta$	solving, inequalities, sign diagrams

The three faces of one quadratic. Most lost marks in this topic are really a student using the wrong face.

If a < 0, every conclusion flips

The parabola opens downwards, the vertex is a maximum, the range is $y \leqslant q$ , and the curve is above the axis between the roots. Papers write $y = 5 + 3x - 2x^2$ precisely because students read the $5$ first and miss the $-2$ . Read the $x^2$ term, not the leading number.

Demonstration — a complete sketch, no calculator, no table of values

Sketch $y = x^2 - 2x - 8$ .

Direction. $a = 1 > 0$ , so it opens upwards; the vertex is a minimum.
$y$ -intercept. $x = 0$ gives $y = -8$ , i.e. the point $(0,-8)$ . That is just $c$ .
Roots. Set $y = 0$ : $x^2 - 2x - 8 = 0$ . Two numbers multiplying to $-8$ and adding to $-2$ are $-4$ and $+2$ , so $(x-4)(x+2) = 0$ and $x = -2 \quad\text{or}\quad x = 4$
Vertex. Midway between the roots: $x = \frac{-2+4}{2} = 1$ — which agrees with $-\frac{b}{2a} = -\frac{-2}{2} = 1$ . Then $y = 1 - 2 - 8 = -9$ , so the vertex is $(1,-9)$ .
Range. The lowest $y$ ever gets is $-9$ , and the curve climbs forever: $y \geqslant -9$

Five facts, no plotting. That is a complete sketch.

02

Completing the square

Syllabus requirement · §1.1

“

carry out the process of completing the square for a quadratic polynomial ax² + bx + c, and use a completed square form.

”

What it is, and why it is worth so much

Completing the square rewrites $ax^2+bx+c$ so that $x$ appears only once, inside a bracket that is squared. Everything the general form hides — vertex, minimum value, range, whether the curve reaches the axis at all — is then printed on the outside.

The argument behind it is one line: a square is never negative. For $y = a(x+p)^2 + q$ with $a > 0$ we have $a(x+p)^2 \geqslant 0$ , so $y \geqslant q$ for every $x$ — and $y = q$ is actually reached, when the bracket is zero, i.e. at $x = -p$ . So $q$ is a genuine floor, not just a bound.

The method, including when a ≠ 1

1
Take $a$ out of the $x^2$ and $x$ terms only. The constant stays outside the bracket. $9x^2 - 36x + 8 \;=\; 9\left(x^2 - 4x\right) + 8$
Dragging the constant inside is the single most common error here.
2
Halve the coefficient of $x$ inside the bracket. Half of $-4$ is $-2$ , so the bracket is $(x-2)^2$ .
3
Subtract the square of that number, inside the bracket. $(x-2)^2 = x^2-4x+4$ is $4$ too big: $9\left[(x-2)^2 - 4\right] + 8$
4
Multiply back out and collect. $9(x-2)^2 - 36 + 8 \;=\; 9(x-2)^2 - 28$
The correction is inside the bracket, so it gets multiplied by a — here −4 becomes −36. This is the classic slip.
5
Check by putting $x = 0$ into both forms: $9(-2)^2 - 28 = 8$ ✓, matching the original constant.

ax^2 + bx + c \;=\; a\left(x + \frac{b}{2a}\right)^{\!2} + \;c - \frac{b^2}{4a}

The general result

·

Worth understanding, not memorising — in the exam, do the four steps on the actual numbers.

Everything the completed square hands you

For $y = a(x+p)^2 + q$ with $a > 0$ :

the vertex is $(-p,\; q)$ , and it is a minimum
the minimum value of $y$ is $q$ , occurring at $x = -p$
the range is $y \geqslant q$
there are no real roots if $q > 0$ (the whole curve sits above the axis)

If $a < 0$ , every one of those flips: maximum, $y \leqslant q$ , no real roots if $q < 0$ .

Completed-square form says exactly how y = x² was moved: (x − 2)² − 3 is “2 right, 3 down”, so the vertex is (2, −3). The signs are opposite in x and the same in y — the trap worth rehearsing.

Match the letters to the form the question printed

Papers ask for the same manipulation under half a dozen labels — $a(x+b)^2+c$ , $p(x+q)^2+r$ , $(x+a)^2+b$ , $-a(x-b)^2+c$ , $a-(x+b)^2$ , $(2x+a)^2+b$ . Only the naming changes, and the marks are for the letters the question asked for.

Form $p(x+q)^2 + r$ , answer $9(x-2)^2 - 28$ ⟹ $q = \mathbf{-2}$ , not $+2$ : the printed bracket has a plus in it, and $(x-2) = (x + (-2))$ .
Form $-a(x-b)^2+c$ with $a$ , $b$ , $c$ positive, answer $-2(x-2)^2+19$ ⟹ $a = \mathbf{2}$ : the minus is already printed, so $a$ supplies only the size.
Form $(2x+a)^2 + b$ ⟹ do not take the $4$ out. Expand the target, $(2x+a)^2+b = 4x^2+4ax+a^2+b$ , and compare coefficients.

The archetype — complete the square, then get paid for it twice

9709/12 O/N 2025 Q15 marks

(a) Express $9x^2 - 36x + 8$ in the form $p(x+q)^2 + r$ , where $p$ , $q$ and $r$ are constants. [2]
(b) Hence find the set of values of the constant $k$ for which the equation $9x^2 - 36x + 8 = k$ has no real roots. [1]
(c) Find the exact roots of the equation $9x^2 - 36x + 8 = -15$ . [2]

Show full working

1
(a) Take the $9$ out of the $x^2$ and $x$ terms only, leaving the $8$ outside: $9x^2 - 36x + 8 = 9\left(x^2 - 4x\right) + 8$
2
Half of $-4$ is $-2$ , and $(x-2)^2 = x^2-4x+4$ is $4$ too big, so $x^2 - 4x = (x-2)^2 - 4$ : $9\Big[(x-2)^2 - 4\Big] + 8$
3
Multiply the $9$ back in — it multiplies the $-4$ too — and collect: $9(x-2)^2 - 36 + 8 = 9(x-2)^2 - 28$
4
Read the letters against the printed form $p(x+q)^2+r$ : $p = 9, \qquad q = -2, \qquad r = -28$
Ours is (x − 2) = (x + (−2)), so q = −2. Writing q = 2 is the commonest way to throw this mark away.
5
(b) $9x^2 - 36x + 8 = k$ asks where the curve meets the horizontal line $y = k$ . From (a) the curve's minimum is $-28$ , so it occupies $y \geqslant -28$ and nothing below. A line under the floor cannot touch it: $k < -28$
6
The sign is strict: at $k = -28$ the line passes through the vertex and there is a repeated root.
The equal case is a real root, not the absence of one — that is the whole mark.
7
(c) Exact forbids decimals, so use the completed square, which has $x$ in only one place: $9(x-2)^2 - 28 = -15 \;\Longrightarrow\; 9(x-2)^2 = 13 \;\Longrightarrow\; (x-2)^2 = \tfrac{13}{9}$
8
Square-root both signs, and simplify: $\sqrt{13/9} = \sqrt{13}/3$ . $x = 2 \pm \frac{\sqrt{13}}{3}$
Stop here. Writing 3.20 and 0.798 loses the mark — 'exact' means surd or fraction form.

Answer

(a) $9(x-2)^2 - 28$ , i.e. $p = 9$ , $q = -2$ , $r = -28$ (b) $k < -28$ (c) $x = 2 \pm \dfrac{\sqrt{13}}{3}$

Part (b) is a 1-mark question that is impossible without (a) and trivial with it. That is the design of the whole topic: the completed square is set up in part (a) and then charged for two or three more times. Never abandon a "hence".

Common mistakes

$9x^2 - 36x + 8 = 9(x-2)^2 - 4 + 8 = 9(x-2)^2 + 4$
$9\left[(x-2)^2 - 4\right] + 8 = 9(x-2)^2 - 28$
The −4 is inside the bracket, so it must be multiplied by the 9 on the way out.
$x^2 + 6x + 1 = (x+3)^2 + 1$
$(x+3)^2 - 9 + 1 = (x+3)^2 - 8$
Squaring the bracket introduces an extra +9 that was never there — take it back out.
$y = (x-2)^2 - 3$ has a minimum at $x = -2$
Minimum at $x = 2$
The bracket is zero when x = 2. The sign inside the bracket is opposite to the position of the vertex.
Stating the range as $x \geqslant 8$
$f(x) \geqslant 8$ (or $y \geqslant 8$ )
A range is a set of output values. The mark scheme for exactly this says 'x ⩾ 8 scores B0' — right working, wrong letter, no mark.
Rounding $\left(x + \tfrac52\right)^2 - \tfrac{21}{4}$ to $(x+2.5)^2 - 5.3$
Keep exact fractions unless decimals are allowed
5.3 ≠ 21/4, so the identity is no longer true. Fractions are the answer, not an unfinished step.

03

Solving a quadratic equation

Syllabus requirement · §1.1

“

solve quadratic equations, and quadratic inequalities, in one unknown.

”

The one fact all of it rests on

If two numbers multiply to give zero, at least one of them is zero.

Nothing else has this property: $AB = 12$ tells you almost nothing, but $AB = 0$ forces $A = 0$ or $B = 0$ . That is why factorising solves an equation — and why it only works once everything is collected on one side and the other side is zero.

Pick the right tool

a(x-\alpha)(x-\beta) = 0

Factorising — fastest, but only when the roots are whole numbers or simple fractions

x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

The formula — always works; the default when factorising fails

a(x+p)^2 + q = 0

Completing the square — when asked for by name, or when the question wants exact surds

The question says…	Use	Because
“Solve”, 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…” / “Use the quadratic formula to…”	Exactly that method	naming a method makes it compulsory — the marks are for the working
“Giving your answers in the form $x = a + b\sqrt{c}$ ”	Completing the square	the printed form is a hint about the shape of the answer
“Correct to 3 significant figures”	The formula, rounding only at the very end	keep full accuracy until the last line

The question almost always tells you which tool it wants. If it names a method, that method is the thing being marked.

Two special cases, and one word that changes everything

No constant term. $3x^2 - 12x = 0$ factorises as $3x(x-4) = 0$ , giving $x = 0$ or $x = 4$ . Do not divide by $x$ — that silently deletes the root $x = 0$ . If the context later forbids $x = 0$ (a length, a number of terms), reject it afterwards, explicitly.

No middle term. $x^2 - 9 = 0$ gives $x = \pm 3$ . Writing only $x = 3$ loses half the answer, and it is the same mistake that wrecks §07.

"Exact" is an instruction. Decimals score zero against it: write $x = 2 \pm \frac{\sqrt{13}}{3}$ , never $3.20$ and $0.798$ . Read the last line of the question before you start writing.

Solving exactly by completing the square

9709/15 M/J 2025 Q3(a)2 marks

Use completing the square to find the exact solutions of the equation $4x^2 - 4x - 1 = 0$ .

Show full working

1
The question names the method, so completing the square is compulsory — the formula would get the right numbers and score zero.
The mark scheme awards the method mark only for producing an (ax + b)² term.
2
Take the $4$ out of the $x^2$ and $x$ terms, leaving the $-1$ outside: $4\left(x^2 - x\right) - 1 = 0$
The coefficient of x inside is −1, which is easy to misread as −4.
3
Half of $-1$ is $-\tfrac12$ , and $\left(x-\tfrac12\right)^2 = x^2 - x + \tfrac14$ overshoots by $\tfrac14$ , so $x^2 - x = \left(x-\tfrac12\right)^2 - \tfrac14$ : $4\left[\left(x - \tfrac12\right)^2 - \tfrac14\right] - 1 = 0$
4
Multiply the $4$ through — note $4 \times \tfrac14 = 1$ — and collect: $4\left(x - \tfrac12\right)^2 - 2 = 0$
5
Undo the square one move at a time: add $2$ , divide by $4$ . $\left(x - \tfrac12\right)^2 = \frac{2}{4} = \frac12$
6
Square-root both sides, with $\pm$ , then add $\tfrac12$ : $x = \frac12 \pm \sqrt{\frac12}$
Both signs. Half the marks in exact-solution questions are lost by taking only the positive root.
7
Tidy if you like: $\sqrt{\tfrac12} = \dfrac{\sqrt2}{2}$ , so $x = \dfrac{1 \pm \sqrt2}{2} = \tfrac12\left(1 \pm \sqrt2\right)$ .

Answer

$x = \dfrac{1}{2} \pm \sqrt{\dfrac{1}{2}}$ , i.e. $x = \tfrac12\left(1 \pm \sqrt2\right)$

Look at the shape: a number $\pm$ a surd over a common denominator. That is what "exact" always produces. If your working hands you $2.207$ and $-0.207$ in a question that said exact, you used the wrong tool.

Common mistakes

$(x-3)(x+5) = 9 \Rightarrow x - 3 = 9$ or $x + 5 = 9$
Expand, collect to zero, then factorise: $x^2+2x-24 = 0 \Rightarrow (x+6)(x-4) = 0$
Splitting a product only works when the product is zero. Nine is not zero.
$3x^2 - 12x = 0 \Rightarrow 3x = 12 \Rightarrow x = 4$
$3x(x-4) = 0 \Rightarrow x = 0$ or $x = 4$
Dividing by x deletes the root x = 0.
In the formula, writing $b^2 = -7^2 = -49$ when $b = -7$
$b^2 = (-7)^2 = +49$
Substitute with brackets round every negative. A negative b² is always an error.
$x^2 = 16 \Rightarrow x = 4$
$x = \pm 4$
Every positive number has two square roots.
Giving $x = 2.78$ and $0.719$ when the question said “exact”
$x = \dfrac{7 \pm \sqrt{17}}{4}$
‘Exact’ forbids decimals outright. Stop the moment you have the surd.

04

Quadratic inequalities

Syllabus requirement · §1.1

“

solve quadratic equations, and quadratic inequalities, in one unknown.

”

The two things that lose the marks

Questions phrased "find the set of values of $x$ for which…" or "the range of possible values of $k$ " are inequalities, and marks leak in two places: students treat $<$ as though it behaved like $=$ , and even with the right two numbers they write the answer down in a notation the mark scheme refuses.

Both are fixed by one habit: find the critical values, look at a sketch, then write the answer as a single object.

The critical values. A continuous curve cannot pass from positive to negative without being zero, so the quadratic can only change sign at its roots. Solve the corresponding equation first — its roots cut the number line into at most three stretches, and on each stretch the sign is constant.

The one algebraic rule. Multiplying or dividing an inequality by a negative number reverses the sign. Avoid ever needing it: arrange the quadratic so the $x^2$ coefficient is positive by moving terms across, not by multiplying by $-1$ .

The only rule you need

For an upward parabola ( $a > 0$ ) with roots $\alpha < \beta$ :

\text{quadratic} < 0 \iff \alpha < x < \beta \qquad\qquad \text{quadratic} > 0 \iff x < \alpha \ \text{ or } \ x > \beta

Less than zero → between. Greater than zero → outside. If $a < 0$ both swap — which is precisely why you rearrange to make $a$ positive first.

The curve is below the axis exactly between its roots and above it outside them. Dropping the roots onto a number line turns the sketch directly into the answer.

Solving a quadratic inequality

1
Get everything to one side, keeping the $x^2$ coefficient positive.
2
Find the critical values — the roots of the corresponding equation.
3
Sketch the parabola. Only the two roots and the direction matter; nothing else on the sketch does.
4
Read off the stretches where the curve is on the side you want.
5
Match the strictness. $<$ or $>$ excludes the endpoints; $\leqslant$ or $\geqslant$ includes them. Copy the question's signs.
6
Write it in one piece. Between the roots is a chain $\alpha < x < \beta$ ; outside is two statements joined by or.
7
Test one value from your answer set in the original inequality.
Three seconds, and it catches the commonest error in the topic — giving the complement of the right answer.

The notation mark schemes actually reject

This is worth more marks than any other paragraph on the page, because examiners write the rule down explicitly.

A "between" answer must be one chain. For $-4 < x < \tfrac23$ the published guidance reads: "Condone $x > -4$ , $x < \tfrac23$ and $x > -4$ and $x < \tfrac23$ — but not $x > -4$ or $x < \tfrac23$ ." The word or turns the statement into "every real number".

An "outside" answer must use or. Writing $0 < k < \tfrac43$ for $k < 0$ or $k > \tfrac43$ is nonsense; the guidance says flatly "Do not accept $0 < k < \tfrac43$ ."

Strict in, strict out. "A0 if $\leqslant$ sign or signs used"; "B0 for use of $\leqslant$ and/or $\geqslant$ ."

Critical values are not an answer. Finding $-5$ and $11$ and stopping is typically one mark of two; the directed inequality is the other.

Reading “decreasing” as an inequality

9709/13 M/J 2025 Q7(a)3 marks

A curve is such that $\dfrac{dy}{dx} = 3x^2 + 10x - 8$ .

Find the set of values of $x$ for which $y$ decreases as $x$ increases.

Show full working

The finished sketch: critical values −4 and 2/3 dropped onto a number line, curve below zero between them — so the answer reads off as one chain.

1
Translate the words. " $y$ decreases as $x$ increases" means the gradient is negative: $\frac{dy}{dx} < 0 \;\Longrightarrow\; 3x^2 + 10x - 8 < 0$
This translation is a mark on its own. Write the inequality down before doing anything with it.
2
The $x^2$ coefficient is already positive and everything is on one side, so go straight to the critical values: solve $3x^2 + 10x - 8 = 0$ .
3
Since $a = 3 \neq 1$ , split the middle term. $a \times c = -24$ and $b = 10$ ; two numbers multiplying to $-24$ and adding to $10$ are $+12$ and $-2$ : $3x^2 + 12x - 2x - 8 = 0$
4
Factorise in pairs, then pull out the common bracket: $3x(x+4) - 2(x+4) = 0 \;\Longrightarrow\; (x+4)(3x-2) = 0$
5
Critical values: $x = -4 \qquad\text{and}\qquad x = \tfrac23$
6
Sketch: upward parabola cutting at $-4$ and $\tfrac23$ . 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.
7
Write it as one chain, smaller value first, with strict signs — at the critical values the gradient is exactly zero, so the curve is flat rather than decreasing: $-4 < x < \frac23$
8
Test: at $x = 0$ (inside), $\frac{dy}{dx} = -8 < 0$ ✓. At $x = 1$ (outside), $\frac{dy}{dx} = 5 > 0$ ✓.

Answer

$-4 < x < \dfrac23$

The published guidance for this exact question: "Condone $x > -4$ , $x < \tfrac23$ … but not $x > -4$ or $x < \tfrac23$ ." One three-letter word is the difference between full marks and losing the last one.

The other case, drawn to scale: above zero (green) outside the roots, below zero (coral) between them. The green pieces drop onto the number line as two separate arrows — hence “or”, never a chain.

Common mistakes

$-x^2 + 4x - 3 > 0 \Rightarrow x^2 - 4x + 3 > 0$
$x^2 - 4x + 3 < 0$
Multiplying an inequality by −1 reverses it. Safer: move everything across instead of multiplying.
$x > 1$ or $x < \frac94$
$1 < x < \frac94$
“Or” makes it true for every real number. A between-the-roots answer is a single chain.
$0 < k < \frac43$ for an “outside the roots” answer
$k < 0$ or $k > \frac43$
Outside means two separate pieces, joined by 'or'. Mark scheme: 'Do not accept 0 < k < 4/3'.
Giving the answer as $x = 1, \frac94$
$1 < x < \frac94$
The critical values are the working; the set of values is the answer.
$(2x-3)^2 < 4 \Rightarrow 2x - 3 < 2$
$-2 < 2x - 3 < 2$
Undoing a square always produces two bounds.
Writing $x \leqslant 2$ when the question used a strict inequality
$x < 2$
Mark schemes say 'A0 if ⩽ sign or signs used'. Copy the question's strictness exactly.

05

The discriminant

Syllabus requirement · §1.1

“

find the discriminant of a quadratic polynomial ax² + bx + c and use the discriminant.

”

What it counts

Inside the quadratic formula everything hangs on the sign of the thing under the square root. That quantity is the discriminant, and it counts the real roots without solving anything. Read it off only once the equation is arranged as $ax^2+bx+c = 0$ .

\Delta = b^2 - 4ac

The discriminant

·

Write it out as its own expression — mark schemes say 'not in the quadratic formula unless b² − 4ac is isolated'.

Discriminant	Real roots	The graph	Wording used in questions
$b^2 - 4ac > 0$	two distinct	crosses the axis twice	“two distinct roots”, “meets at two points”, “intersects twice”
$b^2 - 4ac = 0$	one repeated	touches the axis	“equal roots”, “a repeated root”, “is a tangent to”, “touches”
$b^2 - 4ac < 0$	none	misses the axis entirely	“no real roots”, “does not meet”, “never intersects”
$b^2 - 4ac \geqslant 0$	at least one	crosses or touches	“has real roots”, “the roots are real”

Translate the wording into a symbol before touching any algebra — getting this line wrong scores zero with no recovery. Note that “real roots” includes the repeated case, so it is ⩾ 0, not > 0.

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.

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, set $b^2-4ac = 0$ . Likewise "touches" is $\Delta = 0$ and "does not meet" is $\Delta < 0$ .

Tangency questions almost always also ask for the point of contact, worth two of the marks. Get its $x$ from $x = -\frac{b}{2a}$ of the collected quadratic (the repeated root sits on the axis of symmetry), then find $y$ from the line, not the curve.

Where it is actually used

Almost nobody is asked for the discriminant of a quadratic handed to them. The real question is a line and a curve, with an unknown constant in one of them.

Where a line meets a curve both equations hold at once, so setting them equal and collecting gives one quadratic whose roots are the $x$ -coordinates of the intersections. Counting intersections is therefore counting roots — which 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.

Line meets curve: the five steps

1
Eliminate one variable — usually substitute the line's $y$ into the curve.
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.
3
Write $a$ , $b$ and $c$ down separately, brackets and all.
For 2x² + (k − 3)x + 8 = 0, 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: $\Delta > 0$ , $\Delta = 0$ or $\Delta < 0$ .
5
Solve the resulting inequality in the constant — it is usually itself a quadratic, so find the critical values and choose "inside" or "outside" using §04.

If the unknown sits in the x² coefficient, the equation might not be quadratic

Given $3kx^2 + (k+8)x + 3 = 0$ with "two distinct real roots", the discriminant condition is only half the story: if $k = 0$ the $x^2$ term vanishes and you have the linear equation $8x + 3 = 0$ , with one root, not two. A fully careful answer excludes $k = 0$ . The same caution applies to $cx^2$ , $(2k-3)x^2$ , $\tfrac12k^2x^2$ — all used in the last five years.

Line and curve: the standard four marks

9709/12 F/M 2025 Q14 marks

A curve has equation $y = 5 + 3x - 2x^2$ and a straight line has equation $y = kx + 13$ , where $k$ is a constant.

Find the set of values of $k$ for which the curve and the line do not meet.

Show full working

1
Where they meet the $y$ -values agree, so set the right-hand sides equal: $kx + 13 = 5 + 3x - 2x^2$
2
Collect onto the side that makes the $x^2$ coefficient positive — here the left, since $-2x^2$ becomes $+2x^2$ crossing over: $2x^2 + (k-3)x + 8 = 0$
A positive x² coefficient now means no inequality to flip later. It costs nothing.
3
Write the coefficients out: $a = 2, \qquad b = k - 3, \qquad c = 8$
b is the whole bracket k − 3, not k. This is the most penalised slip in the topic, and writing the line out prevents it.
4
"Do not meet" means no real roots, so $\Delta < 0$ : $(k-3)^2 - 4(2)(8) < 0 \;\Longrightarrow\; (k-3)^2 - 64 < 0 \;\Longrightarrow\; (k-3)^2 < 64$
5
Undo the square two-sidedly — a square is under $64$ exactly when the thing squared lies between $-8$ and $8$ : $-8 < k - 3 < 8$
Writing k − 3 < 8 alone gives k < 11 and silently drops the lower bound.
6
Add $3$ to all three parts: $-5 < k < 11$
7
Check with $k = 0$ (inside): $2x^2 - 3x + 8 = 0$ has $\Delta = 9 - 64 = -55 < 0$ — no intersections ✓.

Answer

$-5 < k < 11$

The published mark scheme awards nothing for using " $> 0$ ", and refuses the final mark for $\leqslant$ signs or an "or" — its guidance reads "CWO. Do not allow 'or'. A0 if $\leqslant$ sign or signs used."

Common mistakes

"Does not meet" $\Rightarrow b^2 - 4ac > 0$
"Does not meet" $\Rightarrow b^2 - 4ac < 0$
Mark schemes score this zero with no recovery. Translate the wording into a symbol first.
$k < -5$ or $k > 11$
$-5 < k < 11$
The quadratic in k opens upwards, so “less than zero” is the region between the critical values.
For $2x^2 + (k-3)x + 8 = 0$ , taking $b = k$
$b = k - 3$ , so $\Delta = (k-3)^2 - 64$
b is the entire coefficient of x after collecting, brackets and all.
For $3kx^2 + (k+8)x + 3 = 0$ , taking $a = 3$
$a = 3k$ — and the equation is only quadratic when $k \neq 0$
The unknown lives in the leading coefficient.
Finding $m = 2$ from $(4+m)^2 = 36$ and stopping
$4 + m = \pm 6$ , so $m = 2$ or $m = -10$
Undoing a square gives two branches. Tangency questions are written so that both exist.
Finding the tangent constant but not the point of contact
Use $x = -\dfrac{b}{2a}$ , then get $y$ from the line
Tangency questions almost always ask for the point too, and it is worth two of the marks.

06

One linear, one quadratic

Syllabus requirement · §1.1

“

solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic.

”

The routine, and why it is everywhere

A point lying on both graphs satisfies both equations, so substituting one into the other gives a single equation in one variable — and its roots are the coordinates of the intersections.

This is the same substitution that opens every discriminant question in §05 and every "where does the line meet the circle" in coordinate geometry. Getting fluent here pays out three times over.

The routine

1
Rearrange the LINEAR equation to make one variable the subject. Pick the cheaper one: if the quadratic contains $y^2$ and $xy$ but no $x^2$ , make $x$ the subject.
2
Substitute into the quadratic, so everything is in one variable.
3
Expand and collect into a three-term quadratic $= 0$ .
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 hand you both partners for one value the way the quadratic can.
6
Check one pair in the original quadratic.

Choosing which variable to eliminate

9709/12 M/J 2025 Q24 marks

Find the coordinates of the points of intersection of the curve and the line with equations $2xy + 5y^2 = 24 \quad \text{and} \quad 2x + y + 4 = 0.$

Show full working

1
Look at the quadratic first: it contains $xy$ and $y^2$ but no $x^2$ , so $x$ appears only to the first power. Eliminating $x$ creates no new squares.
Two seconds of looking. Eliminating y here means substituting a bracket into 5y², which triples the algebra.
2
Make $x$ the subject of the linear equation: $2x = -y - 4 \;\Longrightarrow\; x = \frac{-y-4}{2}$
3
Substitute into $2xy + 5y^2 = 24$ . The $2$ at the front cancels the denominator: $(-y-4)y + 5y^2 = 24$
4
Expand and collect: $-y^2 - 4y + 5y^2 = 24$ , so $4y^2 - 4y - 24 = 0$
5
Divide through by $4$ , then factorise — two numbers multiplying to $-6$ and adding to $-1$ are $-3$ and $+2$ : $y^2 - y - 6 = 0 \;\Longrightarrow\; (y-3)(y+2) = 0 \;\Longrightarrow\; y = 3 \text{ or } y = -2$
6
Find each partner from the linear equation $2x + y + 4 = 0$ . With $y = 3$ : $2x + 7 = 0$ , so $x = -\tfrac72$ . With $y = -2$ : $2x + 2 = 0$ , so $x = -1$ .
7
Present as coordinate pairs, keeping each $y$ with the $x$ that produced it: $\left(-\tfrac72,\ 3\right) \qquad\text{and}\qquad (-1,\ -2)$
8
Check the second pair in the original quadratic: $2(-1)(-2) + 5(-2)^2 = 4 + 20 = 24$ ✓

Answer

$(-1,\ -2)$ and $\left(-\tfrac{7}{2},\ 3\right)$

The question said coordinates, so a list of $x$ - or $y$ -values is not an answer. When a question names its output format — "coordinates", "the set of values", "in the form $p(x+q)^2+r$ " — that phrasing is what the final mark is for. This mark scheme also insists fractions be simplified.

Common mistakes

Solving for $y = 3, -2$ and giving that as the answer
Giving the pairs $\left(-\tfrac72,\ 3\right)$ and $(-1,\ -2)$
Half a coordinate is not a point of intersection.
Substituting the $y$ -values back into the quadratic equation
Substituting back into the linear equation
The quadratic can hand you both partners for one value, so you end up with pairs that are not intersections.
Substituting $y = x + 5$ into $3y^2$ as $3x^2 + 25$
$3(x+5)^2 = 3x^2 + 30x + 75$
(x + 5)² is not x² + 25. Expand the square on its own line before multiplying.
Leaving the answer as $y = -\tfrac{11}{2} + 3$
$y = -\tfrac52$
Mark schemes state 'fractions must be simplified'.

07

Equations quadratic in a function of x

Syllabus requirement · §1.1

“

recognise and solve equations in x which are quadratic in some function of x.

”

The pattern

An equation is quadratic in something else when one power is exactly the square of another: $x^4$ and $x^2$ ; $x^6$ and $x^3$ ; $x$ and $\sqrt{x}$ ; $(2x-3)^4$ and $(2x-3)^2$ ; $\tan^2\theta$ and $\tan\theta$ .

Substitute $u$ for the inner function, solve the ordinary quadratic in $u$ , then come all the way back. The bank's recorded techniques for this sub-topic — "convert back to the original variable", "reject an impossible trigonometric value", "find all angles in the given interval" — are every one of them about the return journey, not the substitution.

The same quadratic wearing different costumes. The shape of the algebra never changes — only what u stands for, and therefore how many values of x or θ each root unpacks into.

Equation	Substitute	Becomes	Going back
$x^4 - 5x^2 + 4 = 0$	$u = x^2$	$u^2 - 5u + 4 = 0$	$x^2 = u$ gives two $x$ for each positive $u$ ; none if $u < 0$
$8x^6 + 215x^3 - 27 = 0$	$u = x^3$	$8u^2 + 215u - 27 = 0$	one real cube root per $u$ , negatives included
$6\sqrt{y} + \dfrac{2}{\sqrt{y}} - 7 = 0$	$u = \sqrt{y}$ (after $\times\sqrt{y}$ )	$6u^2 - 7u + 2 = 0$	$\sqrt{y} \geqslant 0$ , so reject negative $u$ ; then $y = u^2$
$x^3 - 28 + \dfrac{27}{x^3} = 0$	$u = x^3$ (after $\times x^3$ )	$u^2 - 28u + 27 = 0$	one real cube root per $u$
$(2x-3)^2 - \dfrac{4}{(2x-3)^2} - 3 = 0$	$u = (2x-3)^2$	$u^2 - 3u - 4 = 0$	$u$ is a square, so reject negative $u$ ; then $2x-3 = \pm\sqrt{u}$
$\tan^2\theta = 1 + \tan\theta$	$u = \tan\theta$	$u^2 - u - 1 = 0$	any real $u$ ; one angle per $180^\circ$
$8\cos^2\theta - 10\cos\theta + 2 = 0$	$u = \cos\theta$	$8u^2 - 10u + 2 = 0$	need $-1 \leqslant u \leqslant 1$ ; then all angles in the interval
$4\sin^4\theta + 12\sin^2\theta - 7 = 0$	$u = \sin^2\theta$	$4u^2 + 12u - 7 = 0$	$u \geqslant 0$ ; then $\sin\theta = \pm\sqrt{u}$ , then the angles
$2^{2x} - 5(2^x) + 4 = 0$	$u = 2^x$	$u^2 - 5u + 4 = 0$	$u > 0$ always; finish with logs (Paper 2/3)

Every disguise the bank has actually used on Paper 1 in the last five years. The last column is different in every row — that is the part the question is really testing.

Spot, substitute, solve, and come back

1
Spot the pattern — is one power exactly the square of another?
2
Clear fractions first if you need to. $\dfrac{27}{x^3}$ becomes a quadratic in $x^3$ only after multiplying every term by $x^3$ .
Check the thing you multiply by cannot be zero. It never can here — the original expression would be undefined.
3
For a trigonometric equation, get everything into one ratio first — replace $\tan\theta$ by $\frac{\sin\theta}{\cos\theta}$ , or $\sin^2\theta$ by $1-\cos^2\theta$ .
You cannot substitute one u while both sin θ and cos θ are present. This is usually the 'show that' part (a).
4
Substitute $u$ for the inner function and solve the quadratic.
5
Reject the impossible $u$ . A square cannot be negative; $\sqrt{\ }$ cannot be negative; $\sin$ and $\cos$ cannot leave $[-1,1]$ .
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 — a separate equation for every surviving root.
This is the step that gets skipped, and it is where the final marks live.
7
Count. Every solution the interval allows, and no extras outside it.

Getting all the angles, and only those angles

Take the inverse function for the principal value — in $[0^\circ,180^\circ]$ for $\cos^{-1}$ , in $[-90^\circ,90^\circ]$ for $\sin^{-1}$ and $\tan^{-1}$ .
Generate the partners from the graph's symmetry:
- $\cos$ : if $\alpha$ works, so does $360^\circ - \alpha$ , then $\pm 360^\circ$ .
- $\sin$ : if $\alpha$ works, so does $180^\circ - \alpha$ , then $\pm 360^\circ$ .
- $\tan$ : if $\alpha$ works, so does $\alpha + 180^\circ$ , every $180^\circ$ .
Keep only those inside the stated interval — and check both ends of it.

Work in degrees unless the interval is written with $\pi$ in it, and never round an intermediate angle before generating its partners.

Quartic in sine — four angles out

9709/13 O/N 2024 Q44 marks

Solve the equation $4\sin^4\theta + 12\sin^2\theta - 7 = 0$ for $0^\circ \leqslant \theta \leqslant 360^\circ$ .

Show full working

1
$\sin^4\theta = \left(\sin^2\theta\right)^2$ , so this is a quadratic in $\sin^2\theta$ — note, in $\sin^2\theta$ , not in $\sin\theta$ .
There is therefore an extra square root to undo on the way back. That is what produces four angles instead of two.
2
Let $u = \sin^2\theta$ , remembering $0 \leqslant u \leqslant 1$ : $4u^2 + 12u - 7 = 0$
3
Split the middle term: $a \times c = -28$ , $b = 12$ , so $+14$ and $-2$ . $4u^2 + 14u - 2u - 7 = 0 \;\Longrightarrow\; 2u(2u+7) - 1(2u+7) = 0$
4
$(2u+7)(2u-1) = 0 \quad\Longrightarrow\quad u = -\tfrac72 \ \text{ or } \ u = \tfrac12$
5
Reject $u = -\tfrac72$ , since $u = \sin^2\theta$ is a square and cannot be negative.
State the rejection and the reason — it is worth a mark on its own.
6
Go back one level and take both square roots: $\sin^2\theta = \tfrac12 \;\Longrightarrow\; \sin\theta = \pm\frac{1}{\sqrt2}$
This is the step that doubles the number of answers. Keeping only the positive root halves your marks.
7
$\sin\theta = \tfrac{1}{\sqrt2}$ : principal value $45^\circ$ ; sine is positive in quadrants 1 and 2, so also $180^\circ - 45^\circ = 135^\circ$ .
8
$\sin\theta = -\tfrac{1}{\sqrt2}$ : sine is negative in quadrants 3 and 4, related acute angle still $45^\circ$ , so $180^\circ + 45^\circ = 225^\circ$ and $360^\circ - 45^\circ = 315^\circ$ .
9
All four lie inside $0^\circ \leqslant \theta \leqslant 360^\circ$ , and there are no others: $\theta = 45^\circ,\ 135^\circ,\ 225^\circ,\ 315^\circ$
The mark scheme gives one mark for any two correct and the second only for all four with no extras — a missing angle and a spurious one cost the same.

Answer

$\theta = 45^\circ,\ 135^\circ,\ 225^\circ,\ 315^\circ$

Count the layers before you start. $\sin^4 \to \sin^2 \to \sin \to \theta$ is three steps back, and each of the middle two can double the number of answers. A quadratic in $\sin^2\theta$ over a full $360^\circ$ typically gives four angles; a quadratic in $\sin\theta$ gives two.

Common mistakes

$x^4 - 5x^2 + 4 = 0 \Rightarrow x^2 = 1, 4 \Rightarrow x = 1, 2$
$x = \pm 1$ and $x = \pm 2$ — four solutions
Every positive value of x² gives two values of x. The most common error in this sub-topic.
$u = \tfrac13, -\tfrac12$ for $u = \sqrt{x}$ , so $x = \tfrac19$ and $x = \tfrac14$
$\sqrt{x} = -\tfrac12$ is impossible, so only $x = \tfrac19$
Squaring a rejected value silently turns an impossible solution into a plausible-looking one.
Giving $u = 1$ and $u = 27$ as the final answer
$x = 1$ and $x = 3$
The question asked for x. The substitution mark is conditional on returning from it, and is often flagged WWW.
Dropping $\cos\theta = 3$ silently
Write “reject $\cos\theta = 3$ since $|\cos\theta| \leqslant 1$ ”
The rejection is worth a mark only if the reason is visible.
$\sin^2\theta = \tfrac12 \Rightarrow \theta = 45^\circ, 135^\circ$
$\sin\theta = \pm\tfrac{1}{\sqrt2}$ , giving $45^\circ, 135^\circ, 225^\circ, 315^\circ$
Undoing the square produces two equations, not one. Half the angles disappear otherwise.
Reporting $\theta = -11.7^\circ$ because the calculator said so
$\theta = 168.3^\circ$ , adjusted into the given interval
The number of angles is set by the interval, not by the number of roots of the quadratic.

08

Recognising a quadratic in someone else's question

Quadratics is rarely the whole question. On Paper 1 it is the demanding end of a coordinate-geometry, functions, series, trigonometry or differentiation question — more than half the parts tagged Quadratics carry another topic's tag too.

The recognition rule

Whenever a 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 the situation forbids.

Step 4 is where the marks separate. In a progression $n$ must be a positive whole number; in geometry a length must be positive; in a convergent GP $|r| < 1$ ; in a restricted function the root must lie in the domain.

Three signals that a quadratic is coming

"Find the possible values of…" — plural, so the equation ahead has two roots.
"Find the value of the constant $k$ " where $k$ appears in two different conditions — eliminate between them and a quadratic in $k$ appears.
Any phrase about intersections, tangency or "does not meet" — a discriminant, which means 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.

Now do the real questionsReal past-paper questions

Everything on one page

ax^2 + bx + c, \quad a \neq 0

The general quadratic — a ≠ 0 is what makes it one

a\left(x + \tfrac{b}{2a}\right)^{2} + c - \tfrac{b^{2}}{4a}

Completed square form

\text{vertex } \left(-p,\ q\right) \text{ for } y = a(x+p)^2 + q

Turning point, read straight off

a > 0:\ y \geqslant q \qquad a < 0:\ y \leqslant q

Range from the completed square — a statement about y, never about x

x = -\frac{b}{2a}

Axis of symmetry — midway between the roots, and where a repeated root sits

AB = 0 \iff A = 0 \text{ or } B = 0

The zero-product rule — why factorising solves anything

x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

The quadratic formula

\Delta = b^{2} - 4ac

The discriminant

\Delta > 0 \;/\; \Delta = 0 \;/\; \Delta < 0

Two distinct / one repeated (tangent) / no real roots

\alpha < x < \beta \quad\text{vs}\quad x < \alpha \ \text{ or } \ x > \beta

Below zero (between) vs above zero (outside), for a > 0

X^2 < k \iff -\sqrt{k} < X < \sqrt{k}

Undoing a square inside an inequality — two-sided, always

u = x^2,\ x^3,\ \sqrt{x},\ \sin\theta,\ \cos\theta,\ \tan\theta

The disguises — substitute, solve, then come all the way back

Can you do all of these?

Sketch $y = x^2 - 2x - 8$ from nothing: direction, $y$ -intercept, roots, vertex, range
Complete the square for $9x^2 - 36x + 8$ , and give $p$ , $q$ , $r$ for the form $p(x+q)^2 + r$ with the right signs
State the vertex, minimum value and range of $y = 3(x-4)^2 - 7$ — the range in terms of $y$ , not $x$
Solve $3x^2 - 12x = 0$ without dividing by $x$ , and get both roots
Solve $4x^2 - 4x - 1 = 0$ exactly, by completing the square
Solve $2x^2 - 5x - 3 \geqslant 0$ and write the answer with the notation the mark scheme accepts
Explain why $-x^2 + 4x - 3 > 0$ has its solution between the roots
Say what $b^2 - 4ac$ must be for "is a tangent to", "does not meet" and "has real roots"
Turn "the line $y = kx + 13$ does not meet the curve" into an inequality in $k$ and solve it
Given 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 $x^4 - 5x^2 + 4 = 0$ and get all four roots
Solve $4\sin^4\theta + 12\sin^2\theta - 7 = 0$ for $0^\circ \leqslant \theta \leqslant 360^\circ$ and get all four angles
State why $\cos\theta = 3$ must be rejected, in a sentence an examiner would accept