Cambridge IGCSE · Topic 1.18
IGCSE Maths 0580 / Topic 1 / IGCSE

Keep it exact.

Extended. A surd is a root that is not a whole number, such as \(\sqrt{2}\). Instead of rounding it, simplify it by taking out square factors, collect like surds, and clear surds from the bottom of a fraction.

The key idea

Use \(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\) with \(a\) the largest square factor: \(\sqrt{20} = \sqrt{4}\sqrt{5} = 2\sqrt{5}\). To rationalise, multiply top and bottom by \(\sqrt{a}\), or by the conjugate: \((\sqrt{3} - 1)(\sqrt{3} + 1) = 3 - 1 = 2\).

4 2 √20 = 2√5 4² + 2² = 20
Fig. 1: a right-angled triangle with sides 4 and 2 (drawn to scale). Its hypotenuse is exactly √20 = 2√5, about 4.47
Section 01

Simplifying surds.

In the lab, choose Simplify and move the slider through numbers such as 20, 72 and 200. Watch the lab split each number into a square factor times what is left, then take the square root of the square part outside.

Worked example Take out the largest square
Simplify (a) \(\sqrt{72}\) (b) \(\sqrt{75}\).
  1. (a) The largest square factor of 72 is 36: \(\sqrt{72} = \sqrt{36}\sqrt{2}\).
  2. \(\sqrt{36} = 6\), so \(\sqrt{72} = 6\sqrt{2}\).
  3. (b) \(75 = 25 \times 3\), so \(\sqrt{75} = \sqrt{25}\sqrt{3} = 5\sqrt{3}\).
Answer: (a) \(6\sqrt{2}\)   (b) \(5\sqrt{3}\)
Examiner trap

\(\sqrt{72} = 2\sqrt{18}\) is true but NOT fully simplified, because 18 still has the square factor 9. Use the LARGEST square factor, or keep going until no square factor is left.

Section 02

Adding, subtracting and expanding.

Like surds collect like algebra: \(3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\), just as \(3x + 5x = 8x\). Simplify each surd first so you can see which ones are alike. In the lab, try the Add and subtract mode.

Worked example Simplify, then collect
Simplify \(\sqrt{200} - \sqrt{32}\).
  1. \(\sqrt{200} = \sqrt{100}\sqrt{2} = 10\sqrt{2}\).
  2. \(\sqrt{32} = \sqrt{16}\sqrt{2} = 4\sqrt{2}\).
  3. \(10\sqrt{2} - 4\sqrt{2} = 6\sqrt{2}\).
Answer: \(6\sqrt{2}\)
Worked example Expand brackets
Expand and simplify (a) \((2 + \sqrt{3})^{2}\) (b) \((3 + \sqrt{2})(3 - \sqrt{2})\).
  1. (a) \((2 + \sqrt{3})(2 + \sqrt{3}) = 4 + 2\sqrt{3} + 2\sqrt{3} + 3\).
  2. Collect: \(7 + 4\sqrt{3}\).
  3. (b) \(9 - 3\sqrt{2} + 3\sqrt{2} - 2\): the surd terms cancel.
  4. \(= 7\).
Answer: (a) \(7 + 4\sqrt{3}\)   (b) \(7\)
Your turn
Simplify \(\sqrt{12} + \sqrt{27} - \sqrt{48}\).
Examiner trap

\(\sqrt{200} - \sqrt{32}\) is NOT \(\sqrt{168}\). You can only subtract surds once they are the same surd, like \(10\sqrt{2} - 4\sqrt{2}\).

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

Simplify \(\sqrt{50}\).

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Work out \(\sqrt{3} \times \sqrt{12}\).

Quick check+10 XP

Rationalise \(\dfrac{6}{\sqrt{3}}\).

Quick check+10 XP

Simplify \(3\sqrt{2} + 5\sqrt{2}\).

Section 03

Rationalising the denominator.

An answer with a surd on the bottom is not in its simplest form. Multiply the top and the bottom by the same thing, chosen so that the bottom becomes a whole number. The lab's Rationalise mode shows each step.

Worked example A single surd on the bottom
Rationalise the denominator of \(\dfrac{10}{\sqrt{5}}\).
  1. Multiply top and bottom by \(\sqrt{5}\): \(\dfrac{10\sqrt{5}}{\sqrt{5}\sqrt{5}} = \dfrac{10\sqrt{5}}{5}\).
  2. Cancel: \(\dfrac{10}{5} = 2\).
Answer: \(2\sqrt{5}\)
Worked example Use the conjugate
Rationalise the denominator of \(\dfrac{1}{\sqrt{3} - 1}\).
  1. Multiply top and bottom by the conjugate \(\sqrt{3} + 1\) (change the sign in the middle).
  2. Bottom: \((\sqrt{3} - 1)(\sqrt{3} + 1) = 3 + \sqrt{3} - \sqrt{3} - 1 = 2\).
  3. Top: \(1 \times (\sqrt{3} + 1) = \sqrt{3} + 1\).
Answer: \(\dfrac{\sqrt{3} + 1}{2}\)
Worked example Conjugate, then cancel
Write \(\dfrac{4}{3 + \sqrt{5}}\) in the form \(a + b\sqrt{5}\).
  1. Multiply top and bottom by \(3 - \sqrt{5}\).
  2. Bottom: \(9 - 5 = 4\). Top: \(4(3 - \sqrt{5}) = 12 - 4\sqrt{5}\).
  3. Divide by 4: \(3 - \sqrt{5}\).
Answer: \(3 - \sqrt{5}\), so \(a = 3,\ b = -1\)
Examiner trap

Multiplying \(\dfrac{1}{\sqrt{3} - 1}\) by \(\dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}\) does not help: the bottom becomes \(4 - 2\sqrt{3}\), still a surd. Use the conjugate, with the opposite sign.

Stage 2 · Exam

Exam-style questions

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Simplify \(\sqrt{200} - \sqrt{32}\).

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Expand and simplify \((2 + \sqrt{3})^{2}\).

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Rationalise the denominator of \(\dfrac{1}{\sqrt{3} - 1}\).

Exam style+20 XP

Write \(\dfrac{4}{3 + \sqrt{5}}\) in the form \(a + b\sqrt{5}\).

Skill unlocked

Surds, sorted.

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Go deeper · practice
Six original Cambridge-style questions
Simplifying, collecting like surds, expanding brackets, rationalising with a single surd and a conjugate, and an exact triangle.