Algebra

Solving quadratics

9 past-paper questions, easiest first. Try each one, then reveal the answer.

Foundation & Higher2 of 9 from Foundation papers

Revision guide by The Full Marks

Solving quadratics: what you need to know

A quadratic equation has an $x^2$ term and usually two solutions. Factorising is the quickest method; at Higher you also need the quadratic formula and completing the square for ones that won't factorise.

Foundation solves by factorising (with $x^2$ on its own) and from graphs; the quadratic formula, completing the square and harder factorising are Higher.

Key facts

Always rearrange to $ax^2 + bx + c = 0$ first.
If two brackets multiply to 0, one of them must be 0.
Higher: quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Higher: completing the square: $x^2 + bx + c = (x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c$ .
On a graph, the solutions are where the curve crosses the $x$ -axis.

Solving by factorising

Rearrange so one side is 0.
Find two numbers that multiply to $c$ and add to $b$ (when $a = 1$ ).
Write the brackets and set each one equal to 0.

Higher: harder factorising and completing the square

When $a \ne 1$ , e.g. $2x^2 - 3x - 5 = 0$ : find two numbers that multiply to $ac = -10$ and add to $b = -3$ ( $-5$ and $2$ ). Split the middle term and factorise in pairs: $(2x - 5)(x + 1) = 0$ , so $x = 2.5$ or $x = -1$ .
Completing the square: write it as $(x + p)^2 = q$ , then square root both sides, remembering $\pm$ . E.g. $x^2 + 6x - 2 = 0$ gives $(x + 3)^2 = 11$ , so $x = -3 \pm \sqrt{11}$ .

Higher: using the quadratic formula

Write down $a$ , $b$ and $c$ , including their signs.
Substitute into the formula, putting negatives in brackets.
Work out both answers (the $+$ and the $-$ ) and round as asked.

Worked examples

Example 1 Foundation · grade 5

Solve $x^2 + 2x - 15 = 0$ .

Show the solution

Two numbers that multiply to $-15$ and add to $2$ : $5$ and $-3$ .
$(x + 5)(x - 3) = 0$
$x + 5 = 0$ or $x - 3 = 0$
$x = -5$ or $x = 3$

Example 2 Higher · grade 7

Solve $2x^2 - 5x - 4 = 0$ . Give your answers to 2 decimal places.

Show the solution

$a = 2$ , $b = -5$ , $c = -4$
$x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(-4)}}{2(2)} = \frac{5 \pm \sqrt{57}}{4}$
$x = \frac{5 + 7.5498\ldots}{4} = 3.14$
$x = \frac{5 - 7.5498\ldots}{4} = -0.64$

Common mistakes

Getting the signs the wrong way round: $(x + 5)(x - 3) = 0$ gives $x = -5$ and $3$ , not $5$ and $-3$ .
Trying to factorise before rearranging to equal 0.
Forgetting that $-4ac$ with a negative $c$ becomes positive, or dividing only part of the top by $2a$ .
Dividing both sides of $x^2 = 5x$ by $x$ , which loses the solution $x = 0$ . Rearrange to $x^2 - 5x = 0$ and factorise: $x(x - 5) = 0$ .

Exam tips

Write the formula with your numbers substituted before using a calculator: that line usually earns a method mark.
In context questions, reject a negative solution if it can't be a length, and say why.

Questions people ask

How do you solve a quadratic equation?

Rearrange it to equal 0, then factorise into two brackets and set each to 0. If it won't factorise, use the quadratic formula.

When should I use the quadratic formula?

When the question asks for answers to decimal places or significant figures, that's a strong hint it won't factorise, so use the formula.

Exam questions: Solving quadratics (9, easiest first)

Solving quadratics3 marksGrade 5No calculatorFoundation

November 2024 · Paper 1 Foundation · Q26

Solve a quadratic equation by factorising.

Show answer

Answer	$x = 5$ , $x = -3$	3 marks

Worked solution Checked against the mark scheme

Find two numbers that multiply to $-15$ and add to $-2$ : $-5$ and $3$ .

$(x - 5)(x + 3) = 0$

$x = 5$ or $x = -3$

Watch out: Getting the signs the wrong way round: x = −5 and x = 3.

Solving quadratics3 marksGrade 5No calculatorHigher

November 2021 · Paper 1 Higher · Q8

Rearrange a quadratic equation to equal zero and solve it by factorising.

Show answer

Answer	$x = 8$ or $x = -3$	3 marks

Worked solution Checked against the mark scheme

$x^2 - 5x - 24 = 0$

Find two numbers that multiply to $-24$ and add to $-5$ : $-8$ and $3$

$(x - 8)(x + 3) = 0$

$x = 8$ or $x = -3$

Watch out: Getting the signs the wrong way round and writing $-8$ and 3.

Solving quadratics3 marksGrade 5CalculatorFoundation

November 2021 · Paper 2 Foundation · Q27

Show answer

Answer	$x = -2$ or $x = 9$	3 marks

Worked solution Checked against the mark scheme

Find two numbers that multiply to $-18$ and add to $-7$ : $+2$ and $-9$ .

$(x + 2)(x - 9) = 0$

$x = -2$ or $x = 9$

Watch out: Getting the signs the wrong way round: $x = 2$ and $x = -9$ .

Solving quadratics4 marksGrade 6No calculatorHigher

November 2019 · Paper 1 Higher · Q14

Form and solve a quadratic from the area of a right-angled triangle to find its shortest side.

Show answer

Answer	$5$ cm	4 marks

Worked solution Checked against the mark scheme

$\frac{1}{2}(x+4)(x-2) = 27.5$

$(x+4)(x-2) = 55 \Rightarrow x^2 + 2x - 8 = 55$

$x^2 + 2x - 63 = 0 \Rightarrow (x+9)(x-7) = 0$

Lengths are positive, so $x = 7$

Shortest side $= x - 2 = 5$ cm

Watch out: Giving x = 7 as the answer instead of the side length.

Solving quadratics5 marksGrade 7No calculatorHigher

November 2024 · Paper 1 Higher · Q13

Form and solve an equation from the difference in volume of two cuboids with algebraic side lengths.

Show answer

Answer	$x = 7$	5 marks

Worked solution Checked against the mark scheme

Volume A $= (x + 2)(2x - 1)(x - 1) = 2x^3 + x^2 - 5x + 2$

Volume B $= 2x(x + 3)(x - 3) = 2x^3 - 18x$

$2x^3 + x^2 - 5x + 2 = 2x^3 - 18x + 142$

$x^2 + 13x - 140 = 0$ , so $(x + 20)(x - 7) = 0$

Lengths must be positive, so $x = 7$ .

Watch out: Sign errors when expanding the triple product.

Solving quadratics3 marksGrade 7CalculatorHigher

November 2022 · Paper 2 Higher · Q19

Solve a quadratic equation whose leading coefficient is not 1.

Show answer

Answer	$x = -\frac{3}{2}$ or $x = \frac{2}{3}$	3 marks

Worked solution Checked against the mark scheme

Factorise: $6x^2 + 5x - 6 = (2x + 3)(3x - 2)$

$2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$

$3x - 2 = 0 \Rightarrow x = \frac{2}{3}$

Watch out: Sign errors when solving each bracket, e.g. giving 3/2 and −2/3.

Solving quadratics4 marksGrade 7No calculatorHigher

June 2019 · Paper 1 Higher · Q17

Turn a ratio of two algebraic expressions into a quadratic equation and solve it.

Show answer

Answer	$x = -1$ or $x = 2.5$	4 marks

Worked solution Checked against the mark scheme

$\frac{x^2}{3x + 5} = \frac{1}{2}$ , so $2x^2 = 3x + 5$

$2x^2 - 3x - 5 = 0$

$(2x - 5)(x + 1) = 0$

$x = 2.5$ or $x = -1$

Watch out: Cross-multiplying the wrong way to get $x^2 = 2(3x + 5)$ .

Solving quadratics5 marksGrade 8CalculatorHigher

June 2018 · Paper 3 Higher · Q19

Set two tangent ratios equal to form and solve a quadratic equation for x.

Show answer

Answer	$x = \frac{5}{3}$	5 marks

Worked solution Checked against the mark scheme

$\tan e = \frac{x}{4x-1}$ and $\tan f = \frac{6x+5}{12x+31}$

$x(12x+31) = (6x+5)(4x-1)$

$12x^2 + 31x = 24x^2 + 14x - 5$

$12x^2 - 17x - 5 = 0 \Rightarrow (3x-5)(4x+1) = 0$

$x = \frac{5}{3}$ (reject $x = -\frac{1}{4}$ , since $x$ is a length)

Watch out: Giving both roots without rejecting the negative one.

Solving quadratics6 marksGrade 8CalculatorHigher

Specimen · Paper 2 Higher · Q19

Form and solve a quadratic from the area of a right-angled triangle, then use Pythagoras to find its perimeter.

Show answer

Answer	8.64 cm	6 marks

Worked solution Checked against the mark scheme

$\frac{1}{2}x(x - 2) = 2.5 \Rightarrow x^2 - 2x - 5 = 0$

$x = \frac{2 \pm \sqrt{4 + 20}}{2} = 1 \pm \sqrt{6}$ , so $x = 3.449$ (positive)

Sides: 3.449 and 1.449

Hypotenuse $= \sqrt{3.449^2 + 1.449^2} = 3.742$

Perimeter $= 3.449 + 1.449 + 3.742 = 8.64$ cm

Watch out: Forgetting the ½ in the area of a triangle.

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

November 2024 · Paper 2 Higher · Q21: Find the possible values of k for a point R so that the angle at R, between lines to two given points, is a right angle.
June 2024 · Paper 1 Higher · Q21: Use the volume and surface area of a cuboid to form and solve a quadratic for its height.
June 2024 · Paper 2 Higher · Q16: Solve a quadratic inequality given in factorised form.
June 2024 · Paper 2 Higher · Q20: Find the coordinates of the turning point of a quadratic curve.
June 2024 · Paper 2 Foundation · Q24: Complete a table for a quadratic, draw its graph and use it to estimate solutions of an equation.
June 2024 · Paper 3 Higher · Q20: Find unknowns that make three expressions consecutive terms of an arithmetic and then a geometric sequence.
November 2023 · Paper 3 Higher · Q17: Complete the square and state the turning point, solve a quadratic with the formula, and spot an error in factorised solutions.
June 2023 · Paper 1 Higher · Q24: Solve two quadratic inequalities and find the values of x that satisfy both.
June 2023 · Paper 3 Higher · Q21: Solve an equation with two algebraic fractions by clearing denominators and solving the resulting quadratic.
June 2023 · Paper 3 Foundation · Q26: Read the turning point and estimate a root from the graph of a quadratic.
November 2022 · Paper 1 Higher · Q19: Solve an equation involving two algebraic fractions, giving the solutions in surd form.
November 2022 · Paper 2 Foundation · Q24: Read the y-intercept, turning point and approximate roots from the graph of a quadratic.
June 2022 · Paper 1 Higher · Q19: Solve an equation with two algebraic fractions, leading to a quadratic solved in surd form.
June 2022 · Paper 1 Foundation · Q28: Complete a table of values for a quadratic, plot its graph, and read off where it crosses the x-axis.
June 2022 · Paper 2 Higher · Q22: Solve two quadratic inequalities algebraically and give the values of x that satisfy both.
June 2022 · Paper 3 Higher · Q22: Find the intersection points of a straight line and a curve by substituting and solving a quadratic.
November 2021 · Paper 2 Foundation · Q24: Complete a table for a quadratic, draw its graph and use the graph to solve a quadratic equation.
November 2021 · Paper 3 Higher · Q16: Show algebraically that a quadratic curve and a straight line meet at exactly one point.
November 2020 · Paper 1 Higher · Q22: Set the area of a 45° sector ring equal to the area of a circle, form a quadratic and solve for x.
November 2020 · Paper 2 Higher · Q20: Show that a stated probability of a red sweet is impossible with 8 yellow sweets, then use two-sweet probability without replacement to form and solve a quadratic for the number of red sweets.
November 2020 · Paper 3 Higher · Q21: Rearrange a linear equation to find a ratio, then factorise a quadratic in two variables to find another ratio.
November 2019 · Paper 1 Higher · Q18: Show a value of an inverse function, then solve an equation involving a composite function.
November 2019 · Paper 1 Higher · Q21: Sketch a quadratic curve, labelling its turning point and the exact points where it meets the axes.
November 2019 · Paper 2 Higher · Q23: Form a quadratic inequality from the area of a parallelogram using ½ab sin C, then solve it.
November 2019 · Paper 3 Higher · Q17: Draw a suitable straight line on a given quadratic graph to estimate the roots of a different quadratic equation.
June 2019 · Paper 1 Higher · Q19: Complete the square on a quadratic, then use it to state the turning point of its graph.
November 2018 · Paper 1 Higher · Q17: Use two points on a quadratic curve to find its equation, then find its turning point.
November 2018 · Paper 1 Higher · Q22: Form and solve a quadratic equation from a probability of picking two pens of the same colour without replacement.
November 2018 · Paper 3 Higher · Q9: Expand a product of three brackets, find a missing power using index laws, and solve a quadratic with the formula.
November 2018 · Paper 3 Foundation · Q22: Complete a table of values for a quadratic, draw its graph and use it to estimate a root.
June 2018 · Paper 2 Foundation · Q24: Complete a table for a quadratic, plot its graph and use it to estimate solutions of an equation.