Surds [H]
Surds
A surd is a root expression that cannot be simplified to a rational number. In GCSE Mathematics, surds most commonly appear as square roots of integers that are not perfect squares: \( \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7} \) and so on. These values are irrational — their decimal expansions are infinite and non-repeating — so they are always left in root form to preserve exact accuracy.
Not surds (these simplify to rational values): \( \sqrt{4} = 2,quad \sqrt{9} = 3,quad \sqrt{16} = 4,quad \sqrt{25} = 5 \).
A Geometric Example
The diagonal of a square with side length 1 is \( \sqrt{1^2 + 1^2} = \sqrt{2} \) by Pythagoras' theorem. This is an exact, irrational length — no fraction or finite decimal represents it precisely.
Key Surd Laws
| Law | Form | Example |
|---|---|---|
| Product rule | \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \) | \( \sqrt{3} \times \sqrt{12} = \sqrt{36} = 6 \) |
| Quotient rule | \( \dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}} \) | \( \dfrac{\sqrt{50}}{\sqrt{2}} = \sqrt{25} = 5 \) |
| Squaring a surd | \( (\sqrt{a})^2 = a \) | \( (\sqrt{7})^2 = 7 \) |
| Conjugate product | \( (c + \sqrt{d})(c - \sqrt{d}) = c^2 - d \) | \( (\sqrt{5}+2)(\sqrt{5}-2) = 5-4 = 1 \) |
Simplifying Surds
To simplify \( \sqrt{n} \): find the largest perfect square factor of \( n \), split the root using the product rule, and evaluate the perfect square root.
\[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \]The key is choosing the largest perfect square factor — using a smaller one (e.g. 4 for √48) requires an extra step: \( \sqrt{4 \times 12} = 2\sqrt{12} = 2 \times 2\sqrt{3} = 4\sqrt{3} \). Both routes give the same answer, but the largest factor is faster.
Worked Examples
Simplify \( \sqrt{72} \) and \( \sqrt{200} \). Also evaluate \( \sqrt{3} \times \sqrt{27} \).
\[ \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \] \[ \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \] \[ \sqrt{3} \times \sqrt{27} = \sqrt{3 \times 27} = \sqrt{81} = 9 \]Notice that \( \sqrt{3} \times \sqrt{27} \) gives a rational result — this happens when the product under the root is a perfect square.
Simplify \( 5\sqrt{3} + \sqrt{12} - \sqrt{3} \) and \( \sqrt{8} + \sqrt{50} \)
Like surds (same surd factor) can be added or subtracted — treat the surd as a unit, like an algebraic letter.
First simplify any unsimplified surds: \( \sqrt{12} = 2\sqrt{3} \).
\[ 5\sqrt{3} + 2\sqrt{3} - \sqrt{3} = (5 + 2 - 1)\sqrt{3} = 6\sqrt{3} \]Second example: \( \sqrt{8} = 2\sqrt{2},quad \sqrt{50} = 5\sqrt{2} \).
\[ \sqrt{8} + \sqrt{50} = 2\sqrt{2} + 5\sqrt{2} = 7\sqrt{2} \]Rationalise the denominator of \( \dfrac{6}{\sqrt{3}} \) and \( \dfrac{10}{2 + \sqrt{3}} \)
Simple denominator — multiply numerator and denominator by \( \sqrt{3} \):
\[ \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \]Compound denominator — multiply by the conjugate \( (2 - \sqrt{3}) \):
\[ \frac{10}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{10(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{10(2-\sqrt{3})}{4-3} = 10(2-\sqrt{3}) = 20-10\sqrt{3} \]Key Takeaways
- A surd is a root that cannot be simplified to a rational number — it is irrational and must be left in root form to be exact.
- To simplify \( \sqrt{n} \): find the largest perfect square factor of \( n \), split using the product rule, and evaluate the root of the perfect square.
- Only like surds (same surd factor) can be added or subtracted — simplify all surds first before attempting to combine them.
- To rationalise a simple surd denominator \( \sqrt{a} \): multiply top and bottom by \( \sqrt{a} \). For a compound denominator \( (c \pm \sqrt{d}) \): multiply by the conjugate \( (c \mp \sqrt{d}) \).
- The conjugate product \( (c+\sqrt{d})(c-\sqrt{d}) = c^2 - d \) always eliminates the surd from the denominator.