GCSE Maths revision · Higher: number and algebra

Completing the square

Higher only

The idea

Write $x^2 + bx + c$ as $\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$ . For $x^2 + 6x + 2 = (x + 3)^2 - 7$ , the turning point is $(-3, -7)$ .

Halve the coefficient of $x$ to fill the bracket.
Subtract the square of that number.
Add on the constant. The turning point is $(-\frac{b}{2}, \text{the number left})$ .

Watch out Remember to subtract the square you added inside the bracket.

See it worked

Worked exampleNon-calculator3 marks

$y = x^2 + 6x + 2$

(a) Write $x^2 + 6x + 2$ in the form $(x + a)^2 + b$ .

(b) Write down the coordinates of the turning point of $y = x^2 + 6x + 2$ .

(a) Halve the coefficient of $x$ : $a = 3$ . $(x + 3)^2 = x^2 + 6x + 9$
Compensate for the extra 9. $x^2 + 6x + 2 = (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7$
(b) $(x + 3)^2 \ge 0$ and equals 0 when $x = -3$ , so the minimum is $(-3, -7)$ .

Answer (a) $(x + 3)^2 - 7$ ; (b) $(-3, -7)$

Your turn

Question 1Calculator allowed2 marks

Find the coordinates of the turning point of $y = x^{2} + 10x + 17$ .

Question 2Non-calculator4 marks

$y=x^2-6x+10$ .

(a) Write y in completed-square form.

(b) Find the minimum point.

Question 3Non-calculator5 marks

$2x^2 - 12x + 5$

(a) Write $2x^2 - 12x + 5$ in the form $a(x + b)^2 + c$ .

(b) Hence solve $2x^2 - 12x + 5 = 0$ . Give your answers in exact form.

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