Surds

1. Simplify surds like $\sqrt{48}$ by taking out square factors

   Look for the biggest square number hiding inside: 48 = 16 × 3, so $\sqrt{48}$ = $\sqrt{16}$ × $\sqrt{3}$ = $4\sqrt{3}$. Working with the largest square factor gets you there in one step, but pulling out smaller squares repeatedly works too.

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2. Multiply and divide surds, and expand brackets containing surds

   Surds multiply and divide just like whole numbers under one root: $\sqrt{2}$ × $\sqrt{8}$ = $\sqrt{16}$ = 4. Expanding brackets follows normal algebra - $(1+\sqrt{3})(2-\sqrt{3})$ = 2 − $\sqrt{3}$ + $2\sqrt{3}$ − 3 = −1 + $\sqrt{3}$. Collect the rational parts and the surd parts separately.

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3. Rationalise denominators of the form $\frac{a}{\sqrt{b}}$

   A tidy answer never leaves a surd on the bottom. Multiply top and bottom by the surd: $\frac{5}{\sqrt{3}}$ = $\frac{5\sqrt{3}}{3}$, because $\sqrt{3}$ × $\sqrt{3}$ = 3. The value is unchanged - you have multiplied by 1.

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4. Rationalise denominators of the form $\frac{a}{b+\sqrt{c}}$

   Multiply top and bottom by the CONJUGATE - the same expression with the sign flipped. For $\frac{2}{3-\sqrt{5}}$, use $(3+\sqrt{5})$: the bottom becomes 9 − 5 = 4 because the middle terms cancel, leaving $\frac{6+2\sqrt{5}}{4}$ = $\frac{3+\sqrt{5}}{2}$.

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5. Give exact answers in surd form, and know when exact form is asked for

   When a question says "give your answer in exact form" or "in the form $a\sqrt{b}$", a rounded decimal scores nothing - $2\sqrt{3}$ is exact, 3.46 is not. Keep surds through your working rather than reaching for the calculator, and simplify at the end.

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• Taking out a square factor that is not the biggest one and stopping there, so $\sqrt{48}$ is left as $2\sqrt{12}$.Higher only

  Look for the largest square factor: 48 = 16 × 3, so $\sqrt{48}$ = $4\sqrt{3}$. If the number under the root still has a square factor, keep going.

• Combining surds with different numbers under the root, so $\sqrt{2}$ + $\sqrt{3}$ becomes $\sqrt{5}$.Higher only

  Only like surds combine: $3\sqrt{2}$ + $5\sqrt{2}$ = $8\sqrt{2}$, but $\sqrt{2}$ + $\sqrt{3}$ stays as it is. Simplify first, since $\sqrt{8}$ + $\sqrt{2}$ = $2\sqrt{2}$ + $\sqrt{2}$ = $3\sqrt{2}$.

• Multiplying only the bottom by $\sqrt{3}$ when rationalising, so $\frac{5}{\sqrt{3}}$ becomes $\frac{5}{3}$.Higher only

  Multiply the top and the bottom by $\sqrt{3}$, which leaves the value unchanged: $\frac{5}{\sqrt{3}}$ = $\frac{5\sqrt{3}}{3}$, because $\sqrt{3}$ × $\sqrt{3}$ = 3.

• Rationalising $\frac{2}{3-\sqrt{5}}$ by multiplying by 3 − $\sqrt{5}$, which still leaves a surd on the bottom.Higher only

  Multiply the top and the bottom by the conjugate, 3 + $\sqrt{5}$. The bottom becomes 9 − 5 = 4, so the answer is $\frac{6+2\sqrt{5}}{4}$ = $\frac{3+\sqrt{5}}{2}$.

• Giving a rounded decimal such as 3.46 when the question asks for an exact answer.Higher only

  Keep the surd all the way through: $\sqrt{12}$ = $2\sqrt{3}$ exactly, while 3.46 is only close. When a question says exact, the decimal loses the marks.

More Number topics

• Fractions
• Percentages
• Standard Form
• Indices
• Bounds & Rounding
• Factors & Multiples

Keep reading

• How do I revise for GCSE maths?
• What's on the GCSE maths syllabus?
• What are the GCSE Maths grade boundaries?
• The GCSE Maths syllabus, explained