Home / IGCSE Maths 0580 / Algebra and graphs / Solving quadratic equations
Cambridge IGCSE · Topic 2.5
IGCSE Maths 0580 / Topic 2 / IGCSE

Two answers, usually.

Extended. Solve quadratic equations by factorising and by the quadratic formula, give answers exactly in surd form or rounded, and form quadratics from area and fraction problems.

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The key idea

Rearrange to \(ax^{2} + bx + c = 0\) first. If it factorises, set each bracket equal to zero. If not, use \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). The solutions are where the curve \(y = ax^{2} + bx + c\) crosses the \(x\)-axis.

Section 01

Solve by factorising.

In the lab set \(a = 1\), \(b = -5\), \(c = 6\) and look at where the curve crosses the \(x\)-axis: at 2 and 3. Read the working to see the same two answers from factorising and from the formula. Then change \(c\) until the curve no longer reaches the axis and see what happens to \(b^{2} - 4ac\).

Worked example Rearrange, then factorise
Solve \(x^{2} + 3x = 10\).
  1. Make one side zero: \(x^{2} + 3x - 10 = 0\).
  2. Product \(-10\), sum 3: \(5\) and \(-2\), so \((x + 5)(x - 2) = 0\).
  3. \(x + 5 = 0\) or \(x - 2 = 0\).
Answer: \(x = -5\) or \(x = 2\)
Worked example Leading coefficient not 1
Solve \(2x^{2} + 5x - 3 = 0\).
  1. \(ac = -6\); numbers 6 and \(-1\): \(2x^{2} + 6x - x - 3 = 0\).
  2. \(2x(x + 3) - 1(x + 3) = 0\), so \((2x - 1)(x + 3) = 0\).
  3. \(2x - 1 = 0\) gives \(x = \dfrac{1}{2}\); \(x + 3 = 0\) gives \(x = -3\).
Answer: \(x = \dfrac{1}{2}\) or \(x = -3\)
Worked example No constant term
Solve \(x^{2} = 7x\).
  1. \(x^{2} - 7x = 0\).
  2. Common factor: \(x(x - 7) = 0\).
Answer: \(x = 0\) or \(x = 7\)
Your turn
Solve \(3x^{2} - 10x + 8 = 0\).
Examiner trap

Dividing \(x^{2} = 7x\) by \(x\) gives only \(x = 7\) and loses \(x = 0\). Never divide by the unknown; factorise instead.

Section 02

The quadratic formula.

The formula solves every quadratic, whether it factorises or not. It is on the formula page of the exam, but you must substitute carefully.

For \(ax^{2} + bx + c = 0\):

\[x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]

Worked example Answers to 2 decimal places
Solve \(2x^{2} - 3x - 4 = 0\), giving your answers correct to 2 decimal places.
  1. \(a = 2\), \(b = -3\), \(c = -4\).
  2. \(b^{2} - 4ac = 9 - 4(2)(-4) = 9 + 32 = 41\).
  3. \(x = \dfrac{3 \pm \sqrt{41}}{4}\).
  4. \(\dfrac{3 + 6.403...}{4} = 2.350...\) and \(\dfrac{3 - 6.403...}{4} = -0.850...\)
Answer: \(x = 2.35\) or \(x = -0.85\)
Worked example Exact answers in surd form
Solve \(x^{2} + 4x + 1 = 0\), giving your answers in the form \(p \pm \sqrt{q}\).
  1. \(a = 1\), \(b = 4\), \(c = 1\): \(b^{2} - 4ac = 16 - 4 = 12\).
  2. \(x = \dfrac{-4 \pm \sqrt{12}}{2}\).
  3. \(\sqrt{12} = 2\sqrt{3}\), so \(x = \dfrac{-4 \pm 2\sqrt{3}}{2}\).
  4. Divide both terms by 2.
Answer: \(x = -2 \pm \sqrt{3}\)
Your turn
Solve \(3x^{2} + 7x - 2 = 0\), giving your answers correct to 2 decimal places.
Examiner trap

When \(b\) is negative, \(-b\) is positive: for \(b = -3\), \(-b = +3\). And \(-4ac\) with \(c = -4\) becomes \(+32\). Writing every value in brackets prevents both slips.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Solve \((x - 3)(x + 5) = 0\).

Quick check+10 XP

Solve \(x^{2} - 9 = 0\).

Quick check+10 XP

Which is the quadratic formula for \(ax^{2} + bx + c = 0\)?

Quick check+10 XP

Solve \(x^{2} = 7x\).

Section 03

Problems that make quadratics.

Form an equation, rearrange to \(= 0\), solve, then decide which answers make sense. A length or a number of people cannot be negative.

Worked example An area
A rectangle is 3 cm longer than it is wide. Its area is 40 cm2. Find its width.
  1. Width \(x\), length \(x + 3\): \(x(x + 3) = 40\).
  2. \(x^{2} + 3x - 40 = 0\), so \((x + 8)(x - 5) = 0\).
  3. \(x = -8\) or \(x = 5\). A width cannot be negative, so reject \(-8\).
Answer: Width 5 cm (length 8 cm)
Worked example A fractional equation
Solve \(\dfrac{x}{x + 2} = \dfrac{3}{x - 6}\), giving your answers correct to 2 decimal places.
  1. Cross-multiply: \(x(x - 6) = 3(x + 2)\).
  2. \(x^{2} - 6x = 3x + 6\), so \(x^{2} - 9x - 6 = 0\).
  3. Formula: \(b^{2} - 4ac = 81 + 24 = 105\), \(x = \dfrac{9 \pm \sqrt{105}}{2}\).
Answer: \(x = 9.62\) or \(x = -0.62\)
Examiner trap

Do not stop at the solutions of the quadratic. Reject impossible values and answer the question that was asked, which may be a perimeter or a length rather than \(x\).

Stage 2 · Exam

Exam-style questions

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Exam style+20 XP

Solve \(2x^{2} + 5x - 3 = 0\).

Exam style+20 XP

Solve \(2x^{2} - 3x - 4 = 0\), giving your answers correct to 2 decimal places.

Exam style+20 XP

Solve \(x^{2} + 4x + 1 = 0\), giving your answers in exact form.

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A rectangle is 3 cm longer than it is wide and its area is 40 cm2. What is its perimeter?

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Quadratics, solved.

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Go deeper · practice
Six original Cambridge-style questions
Factorising, the formula to 2 decimal places and in surd form, a fractional equation and a garden in context.