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Maths · Further Algebra

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Solving Quadratic Equations

Solving quadratics by factorising, with the formula and by completing the square, and forming them from problems.

  • 9 key terms
  • All boards

Learning Objectives

  1. 1Solve quadratic equations by factorising, including when the coefficient of \(x^2\) is not 1.
  2. 2Link the solutions of a quadratic equation to the roots of its graph.
  3. 3Solve quadratics using the quadratic formula and by completing the square (Higher tier).
  4. 4Form and solve quadratic equations from a problem, and reject answers that do not fit.

Two answers from one equation

A quadratic equation has an \(x^2\) term, and it usually has two solutions. The method that works every time on a non-calculator paper is factorising: get one side to zero, break the quadratic into two brackets, and set each bracket equal to zero. The numbers are always chosen to factorise neatly, so if yours do not, check your arithmetic. Harder questions, such as problems set in a context, use the same method after you have formed the equation yourself.

Solving by factorising

This method rests on one fact: if two things multiply to give zero, one of them must be zero.

  • Step 1: make one side zero

    Rearrange so the equation is \(ax^2 + bx + c = 0\). Solve \(x^2 = 5x - 6\) by rewriting it as \(x^2 - 5x + 6 = 0\).

  • Step 2: factorise

    \(x^2 - 5x + 6 = (x - 2)(x - 3)\).

  • Step 3: set each bracket to zero

    \(x - 2 = 0\) or \(x - 3 = 0\), so \(x = 2\) or \(x = 3\).

  • Step 4: check

    Put each answer back in: \(2^2 - 10 + 6 = 0\) and \(3^2 - 15 + 6 = 0\).

Factorising with a coefficient

Solve \(2x^2 + 7x + 3 = 0\).

Show the solutionHide the solution
  1. 1 Multiply a and c \(2 \times 3 = 6\). Find two numbers that multiply to 6 and add to 7: 6 and 1.
  2. 2 Split the middle term \(2x^2 + 6x + x + 3 = 0\).
  3. 3 Factorise in pairs \(2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0\).
  4. 4 Solve each bracket \(2x + 1 = 0\) gives \(x = -\dfrac{1}{2}\), and \(x + 3 = 0\) gives \(x = -3\).

Answer\(x = -\dfrac{1}{2}\) or \(x = -3\)

Special cases

Some quadratics are quicker to solve without a full factorisation.

  • No x term

    \(x^2 - 49 = 0\) gives \(x^2 = 49\), so \(x = 7\) or \(x = -7\). Never forget the negative root.

  • No constant term

    \(x^2 - 6x = 0\) factorises as \(x(x - 6) = 0\), so \(x = 0\) or \(x = 6\). Do not divide by \(x\), or a solution is lost.

  • Difference of two squares

    \(x^2 - 25 = (x - 5)(x + 5)\), so \(x = \pm 5\).

  • A repeated root

    \(x^2 - 6x + 9 = (x - 3)^2 = 0\) gives just \(x = 3\).

The quadratic formula (Higher tier)

When a quadratic will not factorise easily, the formula always works. On Paper 1 the numbers are chosen so that the square root is exact.

  • The formula

    For \(ax^2 + bx + c = 0\), \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

  • Substitute carefully

    Put brackets round negative numbers, such as \((-3)^2 = 9\).

  • Two answers

    The \(\pm\) gives the two solutions, one with plus and one with minus.

  • Check the number under the root

    If \(b^2 - 4ac\) is negative, there are no real solutions.

Using the formula

(Higher tier) Solve \(2x^2 + 5x - 3 = 0\).

Show the solutionHide the solution
  1. 1 Identify a, b and c \(a = 2\), \(b = 5\), \(c = -3\).
  2. 2 Work out the discriminant \(b^2 - 4ac = 25 - 4 \times 2 \times (-3) = 25 + 24 = 49\).
  3. 3 Substitute \(x = \dfrac{-5 \pm \sqrt{49}}{4} = \dfrac{-5 \pm 7}{4}\).
  4. 4 Two solutions \(x = \dfrac{2}{4} = \dfrac{1}{2}\) or \(x = \dfrac{-12}{4} = -3\).

Answer\(x = \dfrac{1}{2}\) or \(x = -3\)

Completing the square (Higher tier)

Writing a quadratic as a bracket squared plus a number solves it and gives the turning point.

  • The pattern

    \(x^2 + bx = \left(x + \dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2\).

  • Example

    \(x^2 + 6x - 7 = (x + 3)^2 - 9 - 7 = (x + 3)^2 - 16\).

  • Solve

    \((x + 3)^2 = 16\), so \(x + 3 = \pm 4\), giving \(x = 1\) or \(x = -7\).

  • Turning point

    \((x + 3)^2 - 16\) has its minimum at \((-3, -16)\).

Quadratics from problems

Many exam questions build a quadratic from a shape or a number story.

  • Form the equation

    A rectangle with sides \((x + 5)\) and \((x - 2)\) and area 18 gives \((x + 5)(x - 2) = 18\).

  • Expand and rearrange

    \(x^2 + 3x - 10 = 18\), so \(x^2 + 3x - 28 = 0\).

  • Solve

    \((x + 7)(x - 4) = 0\), so \(x = -7\) or \(x = 4\).

  • Reject answers that do not fit

    A length cannot be negative, so \(x = 4\). The sides are 9 cm and 2 cm, and \(9 \times 2 = 18\).

What the OCR exam gives you

OCR prints a formulae sheet with the paper, and the Higher tier sheet includes the quadratic formula.

  • Given at Higher tier

    The formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is on the Higher tier formulae sheet, so you do not have to memorise it, but you must know how to use it.

  • Not given

    Factorising and completing the square are not on the sheet, and the Foundation tier sheet does not have the formula.

Test yourself

  1. 1

    What does \((x - 4)(x + 2) = 0\) tell you?

    Show answerHide answer

    \(x = 4\) or \(x = -2\).

  2. 2

    What are the solutions of \(x^2 - 49 = 0\)?

    Show answerHide answer

    \(x = 7\) or \(x = -7\).

  3. 3

    What is the first step in solving \(x^2 = 3x + 10\)?

    Show answerHide answer

    Rearrange to \(x^2 - 3x - 10 = 0\).

  4. 4

    What is the discriminant \(b^2 - 4ac\) for \(x^2 + 2x - 8\) (Higher tier)?

    Show answerHide answer

    \(4 + 32 = 36\).

  5. 5

    What do the solutions of a quadratic equation look like on its graph?

    Show answerHide answer

    The points where the curve crosses the \(x\)-axis.

Exam technique: solving quadratics

Method marks are easy to collect if the working is laid out.

  • Zero on one side

    Always rearrange first, because factorising only works when one side is 0.

  • Write the brackets

    The factorised form earns a mark even if you slip afterwards.

  • Give both solutions

    Missing the second solution loses the final mark.

  • Check by substituting

    It takes seconds and catches most sign errors.

Summary and exam focus

  • Rearrange to \(ax^2 + bx + c = 0\), factorise, and set each bracket equal to zero.
  • The solutions are the roots of the graph, where it crosses the \(x\)-axis.
  • Remember the negative root in \(x^2 = k\), and never divide by \(x\).
  • The formula and completing the square (Higher tier) solve quadratics that do not factorise easily.

Exam focus

Solve \(x^2 + 5x + 6 = 0\). (3 marks) (3 marks)

Factorise as \((x + 2)(x + 3) = 0\) to get one mark, and then write \(x = -2\) and \(x = -3\) for the others. If you write only one solution, you lose a mark.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Quadratic equation
An equation whose highest power of \(x\) is \(x^2\), usually with two solutions.
Factorise
Write an expression as a product of brackets.
Root
A solution of an equation, shown where its graph crosses the \(x\)-axis.
Discriminant
The expression \(b^2 - 4ac\) from the quadratic formula.
Quadratic formula
A formula that gives the solutions of any quadratic equation.
Completing the square
Rewriting \(x^2 + bx + c\) as \((x + p)^2 + q\).
Repeated root
A root that comes from a bracket squared, such as \((x - 3)^2 = 0\).
Coefficient
The number multiplying a letter, such as 2 in \(2x^2\).
Substitute
Replace a letter with its value.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Solve 3 marks Easier

Solve \(x^2 + x - 12 = 0\).

Mark scheme — 3 marks available

  • \((x + 4)(x - 3)\) — M1
  • \(x = -4\) — A1
  • \(x = 3\) — A1

Model answer

\((x + 4)(x - 3) = 0\), so \(x = -4\) or \(x = 3\).

2. Exam question Solve 3 marks Core

Solve \(x^2 = 10 - 3x\).

Mark scheme — 3 marks available

  • \(x^2 + 3x - 10 = 0\) — M1
  • \((x + 5)(x - 2)\) — M1
  • \(x = -5\) and \(x = 2\) — A1

Model answer

Rearrange to \(x^2 + 3x - 10 = 0\), so \((x + 5)(x - 2) = 0\) and \(x = -5\) or \(x = 2\).

3. Exam question Show that 4 marks Core

The diagram shows a right-angled triangle. The sides are \(x\) cm and \((x + 2)\) cm, and the hypotenuse is 10 cm. (a) Show that \(x^2 + 2x - 48 = 0\). [2 marks] (b) Calculate the length of the shortest side of the triangle. [2 marks]

A right-angled triangle with shorter sides x cm and (x + 2) cm and hypotenuse 10 cm.

Mark scheme — 4 marks available

  • (a) \(x^2 + (x + 2)^2 = 100\) — M1
  • (a) \(x^2 + 2x - 48 = 0\) shown — A1
  • (b) \((x + 8)(x - 6)\) — M1
  • (b) 6 cm — A1

Model answer

(a) By Pythagoras, \(x^2 + (x + 2)^2 = 100\), so \(2x^2 + 4x + 4 = 100\), which gives \(x^2 + 2x - 48 = 0\). (b) \((x + 8)(x - 6) = 0\), so \(x = 6\) as a length is positive. The shortest side is 6 cm.

4. Exam question Solve 3 marks Core

(a) Solve \(x^2 - 81 = 0\). [1 mark] (b) Solve \(x^2 + 9x = 0\). [2 marks]

Mark scheme — 3 marks available

  • (a) \(x = \pm 9\) — B1
  • (b) \(x(x + 9)\) — M1
  • (b) \(x = 0\) and \(x = -9\) — A1

Model answer

(a) \(x = 9\) or \(x = -9\). (b) \(x(x + 9) = 0\), so \(x = 0\) or \(x = -9\).

5. Exam question Solve 3 marks Stretch

Solve \(2x^2 - x - 10 = 0\).

Mark scheme — 3 marks available

  • \((2x - 5)(x + 2)\) — M1
  • \(x = \dfrac{5}{2}\) — A1
  • \(x = -2\) — A1

Model answer

\(2x^2 - 5x + 4x - 10 = x(2x - 5) + 2(2x - 5) = (2x - 5)(x + 2) = 0\), so \(x = \dfrac{5}{2}\) or \(x = -2\).

6. Exam question Solve 3 marks Stretch

Use the quadratic formula to solve \(x^2 + 3x - 5 = 0\). Give your answers in surd form.

Mark scheme — 3 marks available

  • \(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-5)}}{2}\) — M1
  • \(\sqrt{29}\) seen — M1
  • \(\dfrac{-3 \pm \sqrt{29}}{2}\) — A1

Model answer

\(x = \dfrac{-3 \pm \sqrt{9 + 20}}{2} = \dfrac{-3 \pm \sqrt{29}}{2}\).

7. Multiple choice 1 mark Easier

Solve \((x - 2)(x - 3) = 0\).

  1. A \(x = -2\) or \(x = -3\)
  2. B \(x = 2\) or \(x = 3\) Correct
  3. C \(x = 2\) or \(x = -3\)
  4. D \(x = 6\)

Why: Each bracket can be zero: \(x - 2 = 0\) or \(x - 3 = 0\).

8. Multiple choice 1 mark Easier

Solve \(x^2 - 49 = 0\).

  1. A \(x = 7\) or \(x = -7\) Correct
  2. B \(x = 7\) only
  3. C \(x = 49\)
  4. D \(x = 24.5\)

Why: \(x^2 = 49\) has two square roots.

9. Multiple choice 1 mark Core

Solve \(x^2 - 6x = 0\).

  1. A \(x = 6\) only
  2. B \(x = 0\) or \(x = -6\)
  3. C \(x = 3\)
  4. D \(x = 0\) or \(x = 6\) Correct

Why: \(x(x - 6) = 0\), so \(x = 0\) or \(x = 6\).

10. Multiple choice 1 mark Easier

Factorise \(x^2 - 5x + 6\).

  1. A \((x + 2)(x + 3)\)
  2. B \((x - 1)(x - 6)\)
  3. C \((x - 2)(x - 3)\) Correct
  4. D \((x + 2)(x - 3)\)

Why: The numbers multiply to 6 and add to \(-5\): \(-2\) and \(-3\).

11. Multiple choice 1 mark Core

What is the first step in solving \(x^2 = 3x + 10\) by factorising?

  1. A Divide both sides by \(x\)
  2. B Rearrange to \(x^2 - 3x - 10 = 0\) Correct
  3. C Take the square root of both sides
  4. D Factorise the right-hand side

Why: One side must be zero before you factorise.

12. Multiple choice 1 mark Core

Solve \(x^2 + 5x + 6 = 0\).

  1. A \(x = -2\) or \(x = -3\) Correct
  2. B \(x = 2\) or \(x = 3\)
  3. C \(x = -1\) or \(x = -6\)
  4. D \(x = 5\) or \(x = 6\)

Why: \((x + 2)(x + 3) = 0\).

13. Multiple choice 1 mark Stretch

Solve \(2x^2 + 7x + 3 = 0\).

  1. A \(x = \dfrac{1}{2}\) or \(x = 3\)
  2. B \(x = -2\) or \(x = -3\)
  3. C \(x = -\dfrac{1}{3}\) or \(x = -2\)
  4. D \(x = -\dfrac{1}{2}\) or \(x = -3\) Correct

Why: \((2x + 1)(x + 3) = 0\).

14. Multiple choice 1 mark Stretch

What is the value of \(b^2 - 4ac\) for \(x^2 + 2x - 8 = 0\)?

  1. A \(-28\)
  2. B \(32\)
  3. C \(36\) Correct
  4. D \(4\)

Why: \(4 - 4 \times 1 \times (-8) = 4 + 32 = 36\).

15. Multiple choice 1 mark Stretch

Use the quadratic formula to solve \(2x^2 + 5x - 3 = 0\).

  1. A \(x = -\dfrac{1}{2}\) or \(x = 3\)
  2. B \(x = \dfrac{1}{2}\) or \(x = -3\) Correct
  3. C \(x = \dfrac{1}{2}\) or \(x = 3\)
  4. D \(x = 2\) or \(x = -3\)

Why: \(x = \dfrac{-5 \pm \sqrt{49}}{4} = \dfrac{-5 \pm 7}{4}\).