The function $f$ is defined by $f(x) = x^2 - 4x + 8$ for $x \in \mathbb{R}$. (i) Express $x^2 - 4x + 8$ in the form $(x - a)^2 + b$. (ii) Hence find the set of values of $x$ for which $f(x) < 9$, giving your answer in exact form.

(i) (x - 2)^2 + 4 · (ii) 2 - √5 < x < 2 + √5

Mathematics 9709/13 — May/June 2019

Cambridge AS Level · Pure Mathematics 1 · worked solutions for every part, with the mark scheme

questions

marks

105

minutes

Topics Quadratics · Trigonometry · Coordinate Geometry · Series · Functions · Differentiation · +2 more

Q1QuadraticsFree sample

The function $f$ is defined by $f(x) = x^2 - 4x + 8$ for $x \in \mathbb{R}$ .

(i)

Express $x^2 - 4x + 8$ in the form $(x - a)^2 + b$ .

DifficultyEasy

Worked solution

Approach

Halve the coefficient of $x$ to form the bracket $(x - a)^2$ , then adjust the constant term so the expression is unchanged.

Working

x^2 - 4x + 8 = (x - 2)^2 - 4 + 8

= (x - 2)^2 + 4

So $a = 2$ and $b = 4$ .

Answer

(x - 2)^2 + 4

Final answer

(x - 2)^2 + 4

Detailed explanation

Walkthrough

We want to write $x^2 - 4x + 8$ as $(x - a)^2 + b$ . Take the coefficient of $x$ , which is $-4$ , halve it to get $-2$ , and write $(x - 2)^2$ . Expanding gives $x^2 - 4x + 4$ . Since the original expression has $+8$ , we add the difference, $8 - 4 = 4$ , to get $(x - 2)^2 + 4$ . Hence $a = 2$ and $b = 4$ .

Key Takeaways

Completing the square rewrites a quadratic as a perfect square plus a constant. It is used later for finding turning points and solving inequalities. The procedure: halve the $x$ -coefficient, put it in the bracket, subtract its square, and keep the original constant.

Common Mistakes

Forgetting to subtract the square of half the coefficient, e.g. writing $(x - 2)^2 + 8$ instead of $(x - 2)^2 + 4$ .
Sign error on $a$ , writing $a = -2$ instead of $a = 2$ .

Things to Be Careful About

The value $b$ is the constant after completing the square, not the original constant $8$ .
The mark scheme awards B1 for $(x-2)^2$ and a second B1, dependent on the first, for the $+4$ .

Techniques used

complete the square

(ii)

Hence find the set of values of $x$ for which $f(x) < 9$ , giving your answer in exact form.

DifficultyMedium-Easy

Worked solution

Approach

Use the completed-square form from part (i), substitute into $f(x) < 9$ , rearrange to isolate $(x - 2)^2$ , then take square roots to obtain the exact interval for $x$ .

Working

From part (i), $f(x) = (x - 2)^2 + 4$ . We require $f(x) < 9$ :

(x - 2)^2 + 4 < 9

(x - 2)^2 < 5

Taking square roots of both sides:

-\sqrt{5} < x - 2 < \sqrt{5}

Adding $2$ throughout:

2 - \sqrt{5} < x < 2 + \sqrt{5}

Answer

2 - \sqrt{5} < x < 2 + \sqrt{5}

Final answer

2 - √5 < x < 2 + √5

Detailed explanation

Walkthrough

From part (i) we have $f(x) = (x-2)^2 + 4$ . The condition $f(x) < 9$ becomes $(x-2)^2 + 4 < 9$ . Subtract $4$ from both sides to get $(x-2)^2 < 5$ . For a square to be less than $5$ , the quantity inside must lie strictly between $-\sqrt{5}$ and $\sqrt{5}$ : this gives $-\sqrt{5} < x - 2 < \sqrt{5}$ . Finally add $2$ to every part of the inequality to isolate $x$ , producing $2 - \sqrt{5} < x < 2 + \sqrt{5}$ . This is the exact set of values requested.

Key Takeaways

To solve a quadratic inequality of the form $(x - c)^2 < k$ with $k > 0$ , write $-\sqrt{k} < x - c < \sqrt{k}$ and then solve for $x$ . The answer must be given in exact surd form.

Common Mistakes

Only keeping the upper bound: writing $x - 2 < \sqrt{5}$ without the lower bound $-\sqrt{5} < x - 2$ .
Not using the 'hence': the mark scheme requires using part (i); a non-hence method that is completely correct scores only SC 1/3.
Writing $\leq$ instead of $<$ (though the mark scheme condones $\leq$ ).

Things to Be Careful About

The final answer must contain two inequalities for $x$ , e.g. two separate statements or one combined statement.
The final answer must be exact: $\sqrt{5}$ should not be replaced by its decimal in the answer, although decimals are allowed for the method.
The mark scheme permits $𝑓$ from their part (i) — if a candidate clearly slips and obtains, e.g. $\sqrt{13}$ instead of $\sqrt{5}$ , this is accepted for the method mark.

Techniques used

apply the completed-square formsolve a quadratic inequalityextract the exact interval for x

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