GCSE Maths revision · Higher: number and algebra

The quadratic formula

Higher only

The idea

Any quadratic $ax^2 + bx + c = 0$ can be solved with $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Use it when the quadratic does not factorise.

Rearrange to $ax^2 + bx + c = 0$ and write down $a$ , $b$ and $c$ with their signs.
Substitute, putting negative values in brackets.
Work out both answers (if $b^2 - 4ac < 0$ there are none) and round as the question asks.

Watch out The whole top, $-b \pm \sqrt{\dots}$ , is divided by $2a$ .

Worked exampleCalculator allowed3 marks

Solve $2x^2 + 5x - 4 = 0$ . Give each solution correct to 2 decimal places.

(a) Write down the larger solution.

(b) Write down the smaller solution.

(a) Use the quadratic formula with $a = 2$ , $b = 5$ , $c = -4$ . $x = \frac{-5 \pm \sqrt{25 + 32}}{4} = \frac{-5 \pm \sqrt{57}}{4}$
$x = \frac{-5 + 7.5498\ldots}{4} = 0.6374\ldots \approx 0.64$
(b) $x = \frac{-5 - 7.5498\ldots}{4} = -3.1374\ldots \approx -3.14$

Answer (a) $0.64$ ; (b) $-3.14$

Question 1Non-calculator5 marks

Solve $2x^2+2x-5=0$ .

(a) Find both x values exactly.

(b) Find the product of the two x values.

Write your answer, then show the working

(a) Calculate the discriminant using the signed constant. $2^{2}-4\times 2\times (-5)$
Use both signs over the full denominator 2a. $\frac{-2+\sqrt{44}}{4}$
Therefore $\frac{-2\pm\sqrt{44}}{4}$ .
(b) Multiply the conjugate numerators over the squared denominator. $\frac{4-44}{16}$
Therefore $-\frac{5}{2}$ .

Answer (a) $\frac{-2\pm\sqrt{44}}{4}$ ; (b) $-\frac{5}{2}$

Question 2Calculator allowed5 marks

A rectangle has width x cm and length $(2x+3)$ cm. Its area is 6 cm².

(a) Find the width to 3 significant figures.

(b) Find the length to 3 significant figures.

Question 3Non-calculator4 marks

Solve $\dfrac{x + 1}{x} = 3x - 2$ . Give your solutions in exact form.

Solve the equation. You must show your working.

27 official questions in 27 papers since 2017. The latest:

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