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Cambridge IGCSE Mathematics — 0580 Extended

Topic 1.15: Number — Surds

Simplifying surds

A surd is an irrational root left in exact form, usually \(\sqrt{n}\) where \(n\) is not a perfect square.

Factor the number under the root and take out perfect squares:

\[\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}\]

Method

  1. Find the largest square factor of the number under the root.
  2. Write \(\sqrt{ab} = \sqrt{a}\sqrt{b}\) where \(\sqrt{a}\) is an integer.
  3. Leave the remaining surd in simplest form.

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

Worked solution: square root 200 minus square root 32 equals 6 root 2
Simplify first, then subtract like surds: \(10\sqrt{2} - 4\sqrt{2} = 6\sqrt{2}\)

Operations with surds

Adding/subtracting: only like surds combine — same number under the root.

\[3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}\]

\(3\sqrt{5} + 2\sqrt{3}\) cannot be simplified (unlike surds).

Multiplying: \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\). Simplify the result.

Division: \(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}\) — often rationalise the denominator next.

Simplify \(2\sqrt{3} \times 4\sqrt{12}\).

Worked solution: 2 root 3 times 4 root 12 equals 48
Multiply then simplify: \(8\sqrt{36} = 48\)

Rationalising the denominator

Exam answers should not have a surd in the denominator. Multiply top and bottom by a suitable surd to remove it.

Which method?

Denominator is \(\sqrt{a}\) alone

Multiply top and bottom by \(\sqrt{a}\).

Denominator is \(a \pm \sqrt{b}\)

Multiply top and bottom by the conjugate \(a \mp \sqrt{b}\).

Rationalise \(\dfrac{10}{\sqrt{5}}\).

Worked solution: 10 over root 5 rationalises to 2 root 5
Multiply by \(\sqrt{5}/\sqrt{5}\): \(\dfrac{10}{\sqrt{5}} = 2\sqrt{5}\)

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

Worked solution: 1 over root 3 minus 1 rationalises to root 3 plus 1 over 2
Multiply by the conjugate: \(\dfrac{\sqrt{3}+1}{2}\)

Paper 2 surd skills

Paper 2 (non-calculator)

Know square numbers up to 144 to simplify quickly: \(\sqrt{72} = 6\sqrt{2}\), \(\sqrt{50} = 5\sqrt{2}\). For rationalising, spot when the denominator is a difference of squares: \((\sqrt{3}-1)(\sqrt{3}+1) = 2\).

Try this

Simplify \(\sqrt{48} + \sqrt{27}\).

Show answer
Answer

  1. Take out the largest square factor of each surd

    \[ \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \]
    \[ \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \]

  2. Add like surds

    \[ 4\sqrt{3} + 3\sqrt{3} = 7\sqrt{3} \]

  3. Final answer

    \[ 7\sqrt{3} \]

Exam Traps

  • \(3\sqrt{5} + 2\sqrt{3}\) cannot be added — only like surds (same number under the root) combine.
  • For a denominator \(a \pm \sqrt{b}\), multiply by the conjugate. Multiplying by \(\sqrt{b}\) alone does not clear it.

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