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Maths · Equations and inequalities

Solving quadratic equations 2

Solve harder quadratics by factorising when the coefficient of x squared is not 1, and use the quadratic formula when they do not factorise.

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  • 6 key terms
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Teacher resources

The teacher copies: slides with the questions built in, the answers, and anything else attached to this lesson for whoever is teaching it.

Student handouts

The same files the students see, to print or hand out.

Warm-up

Answer each one, then check.

  1. 1

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

    Show answerHide answer

    \(x = 2\) or \(x = 3\)

  2. 2

    Expand \((2x + 1)(x + 3)\).

    Show answerHide answer

    \(2x^2 + 7x + 3\)

  3. 3

    Work out \(\sqrt{81}\).

    Show answerHide answer

    9

  4. 4

    Work out \(5^2 - 4 \times 1 \times 6\).

    Show answerHide answer

    1

  5. 5

    Round 5.385 to 2 decimal places.

    Show answerHide answer

    5.39

Learning Objectives

  1. 1Factorise and solve \(ax^2 + bx + c = 0\).
  2. 2Use the quadratic formula.
  3. 3Give answers to a given accuracy or as surds.
  4. 4Decide which method to use.

Factorising \(ax^2 + bx + c\)

Use the product \(ac\) and the sum \(b\).

  1. 1 Multiply \(a\) by \(c\)

    For \(2x^2 + 7x + 3\): \(2 \times 3 = 6\)
  2. 2 Find two numbers with that product and sum \(b\)

    1 and 6 (they add to 7)
  3. 3 Split the middle term

    \(2x^2 + x + 6x + 3\)
  4. 4 Factorise in pairs

    \(x(2x + 1) + 3(2x + 1)\)
  5. 5 Write as two brackets

    \((2x + 1)(x + 3)\)

A Harder Factorisation

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

Show the solutionHide the solution
  1. 1 Factorise \((2x + 1)(x + 3) = 0\)
  2. 2 Set each bracket to zero \(2x + 1 = 0\) or \(x + 3 = 0\)
  3. 3 Solve \(x = -\dfrac{1}{2}\) or \(x = -3\)

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

Another Factorisation

Solve \(3x^2 - 10x + 8 = 0\).

Show the solutionHide the solution
  1. 1 \(ac = 24\); two numbers with product 24 and sum \(-10\) \(-4\) and \(-6\)
  2. 2 Split and factorise \(3x^2 - 4x - 6x + 8 = x(3x - 4) - 2(3x - 4)\)
  3. 3 Write as brackets \((3x - 4)(x - 2) = 0\)
  4. 4 Solve \(x = \dfrac{4}{3}\) or \(x = 2\)

Answer\(x = \dfrac{4}{3}\) or \(x = 2\)

THE QUADRATIC FORMULA

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

Learn it: it is not on the formulae sheet.

Using the Formula

Solve \(x^2 + 3x - 5 = 0\), giving your answers to 2 decimal places.

Show the solutionHide the solution
  1. 1 Identify \(a = 1\), \(b = 3\), \(c = -5\) Substitute into the formula
  2. 2 \(b^2 - 4ac\) \(9 - 4(1)(-5) = 9 + 20 = 29\)
  3. 3 Write the formula \(x = \dfrac{-3 \pm \sqrt{29}}{2}\)
  4. 4 Calculate \(\sqrt{29} = 5.385...\)
  5. 5 The two solutions \(x = 1.19\) or \(x = -4.19\)

Answer\(x = 1.19\) or \(x = -4.19\)

Leaving the Answer as a Surd

Solve \(x^2 - 6x + 2 = 0\), giving your answers in the form \(p \pm \sqrt{q}\).

Show the solutionHide the solution
  1. 1 \(a = 1\), \(b = -6\), \(c = 2\) \(b^2 - 4ac = 36 - 8 = 28\)
  2. 2 Formula \(x = \dfrac{6 \pm \sqrt{28}}{2}\)
  3. 3 Simplify \(\sqrt{28} = 2\sqrt{7}\) \(x = \dfrac{6 \pm 2\sqrt{7}}{2}\)
  4. 4 Divide through by 2 \(x = 3 \pm \sqrt{7}\)

Answer\(x = 3 + \sqrt{7}\) or \(x = 3 - \sqrt{7}\)

Which Method?

Factorising

  • Quick when the numbers are simple.
  • Works when the roots are rational.
  • Gives exact answers.

The quadratic formula

  • Always works.
  • Needed for "give to 2 decimal places" or surds.
  • Take care with the signs of \(b\) and \(c\).

Common Mistakes

The formula has plenty of places to slip.

  • Sign of b

    The formula starts with \(-b\); if \(b = -6\), then \(-b = 6\).

  • The bracket

    \(b^2\) must be squared before any minus: \((-6)^2 = 36\).

  • Dividing

    Divide the WHOLE numerator by \(2a\), not just the surd.

  • Order of the calculation

    Use brackets on the calculator for the top line.

Choose Your Method

For each equation, decide whether to factorise or use the formula, then solve it. (a) \(2x^2 - 5x - 12 = 0\) (b) \(x^2 + 4x - 3 = 0\) (to 2 d.p.) (c) \(x^2 - 10x + 25 = 0\).

1. Test for simple factors.

2. Otherwise use the formula.

A good answer shows: (a) \((2x + 3)(x - 4) = 0\): \(x = -1.5\) or 4. (b) Formula: \(\dfrac{-4 \pm \sqrt{28}}{2} = -2 \pm \sqrt{7}\): \(x = 0.65\) or \(-4.65\). (c) \((x - 5)^2 = 0\): \(x = 5\) (one repeated solution).

Can I...?

  1. 1Factorise \(ax^2 + bx + c\).
  2. 2Solve by factorising.
  3. 3Write down the quadratic formula.
  4. 4Identify \(a\), \(b\) and \(c\) correctly.
  5. 5Substitute carefully with brackets.
  6. 6Give answers to 2 decimal places.
  7. 7Give answers as surds.
  8. 8Choose between methods.

Summary & Exam Focus

  • \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
  • Factorise when the numbers are simple; otherwise use the formula.
  • Watch the signs and divide the whole numerator by \(2a\).
  • Check by substituting your answers into the original equation.

Exam focus

Solve \(x^2 + 3x - 5 = 0\). Give your solutions correct to 2 decimal places. (3 marks) (3 marks)

Write \(a\), \(b\) and \(c\) first, and put brackets round negative numbers when you substitute.

Key terms

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

Coefficient
The number multiplying a variable, such as the 2 in \(2x^2\).
Quadratic formula
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Discriminant
The expression \(b^2 - 4ac\) under the square root.
Surd
A root that cannot be simplified to a whole number, such as \(\sqrt{7}\).
Root
A solution of an equation.
Substitute
Replace letters with numbers.

Questions and answers

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

1. Exam question Non-calculator 3 marks Easier

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

Mark scheme — 3 marks available

  • Correct factorisation — M1
  • One solution — A1
  • Both solutions — A1

Model answer

\((2x + 1)(x + 3) = 0\), so \(x = -\dfrac{1}{2}\) or \(x = -3\).

2. Exam question Non-calculator 3 marks Easier

Solve \(3x^2 - 10x + 8 = 0\).

Mark scheme — 3 marks available

  • Correct factorisation — M1
  • One solution — A1
  • Both solutions — A1

Model answer

\((3x - 4)(x - 2) = 0\), so \(x = \dfrac{4}{3}\) or \(x = 2\).

3. Exam question Non-calculator 4 marks Easier

Solve \(2x^2 - 5x - 12 = 0\).

Mark scheme — 4 marks available

  • \(2x^2 - 8x + 3x - 12\) — M1
  • \((2x + 3)(x - 4)\) — M1
  • One solution — A1
  • Both solutions — A1

Model answer

\((2x + 3)(x - 4) = 0\), so \(x = -\dfrac{3}{2}\) or \(x = 4\).

4. Exam question Calculator 3 marks Easier

Solve \(x^2 + 3x - 5 = 0\). Give your solutions correct to 2 decimal places.

Mark scheme — 3 marks available

  • Correct substitution into the formula — M1
  • \(\dfrac{-3 \pm \sqrt{29}}{2}\) — A1
  • \(1.19\) and \(-4.19\) — A1

Model answer

\(x = \dfrac{-3 \pm \sqrt{9 + 20}}{2} = \dfrac{-3 \pm \sqrt{29}}{2}\), so \(x = 1.19\) or \(x = -4.19\).

5. Exam question Calculator 3 marks Easier

Solve \(3x^2 - 4x - 2 = 0\). Give your solutions correct to 2 decimal places.

Mark scheme — 3 marks available

  • Correct substitution — M1
  • \(\dfrac{4 \pm \sqrt{40}}{6}\) — A1
  • \(1.72\) and \(-0.39\) — A1

Model answer

\(x = \dfrac{4 \pm \sqrt{16 + 24}}{6} = \dfrac{4 \pm \sqrt{40}}{6}\), so \(x = 1.72\) or \(x = -0.39\).

6. Exam question Non-calculator 4 marks Easier

Solve \(x^2 - 6x + 2 = 0\). Give your answers in the form \(p \pm \sqrt{q}\), where \(p\) and \(q\) are integers.

Mark scheme — 4 marks available

  • Correct substitution — M1
  • \(\sqrt{28}\) — A1
  • \(2\sqrt{7}\) — M1
  • \(3 \pm \sqrt{7}\) — A1

Model answer

\(x = \dfrac{6 \pm \sqrt{36 - 8}}{2} = \dfrac{6 \pm \sqrt{28}}{2} = \dfrac{6 \pm 2\sqrt{7}}{2} = 3 \pm \sqrt{7}\).

7. Multiple choice 1 mark Easier

In \(3x^2 + 5x - 2 = 0\), what is \(a\)?

  1. A 3 Correct
  2. B 5
  3. C \(-2\)
  4. D 1

Why: \(a\) is the coefficient of \(x^2\): 3.

8. Multiple choice 1 mark Core

What is \(b^2 - 4ac\) for \(x^2 + 3x - 5 = 0\)?

  1. A \(-11\)
  2. B 9
  3. C 29 Correct
  4. D \(-29\)

Why: \(9 - 4(1)(-5) = 9 + 20 = 29\).

9. Multiple choice 1 mark Core

Factorise \(2x^2 + 7x + 3\).

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

Why: \((2x + 1)(x + 3)\) expands to \(2x^2 + 7x + 3\).

10. Multiple choice 1 mark Core

The formula gives \(x = \dfrac{6 \pm \sqrt{28}}{2}\). What is this in simplest form?

  1. A \(3 \pm \sqrt{28}\)
  2. B \(6 \pm \sqrt{14}\)
  3. C \(3 \pm 2\sqrt{7}\)
  4. D \(3 \pm \sqrt{7}\) Correct

Why: \(\sqrt{28} = 2\sqrt{7}\), so divide through by 2: \(3 \pm \sqrt{7}\).

11. Multiple choice 1 mark Core

When is the quadratic formula most useful?

  1. A When the roots are not whole numbers Correct
  2. B When there is no x term
  3. C When the equation is linear
  4. D When c is zero

Why: When the equation will not factorise easily, especially if the answer must be to 2 decimal places.

12. Multiple choice 1 mark Stretch

The equation \(x^2 - 10x + 25 = 0\) has...

  1. A Two different solutions
  2. B One repeated solution Correct
  3. C No solutions
  4. D Three solutions

Why: It is \((x - 5)^2 = 0\), one repeated solution \(x = 5\).