Manipulating surds

Why surds are in a complex numbers standard

• A surd is a root that cannot be written exactly as a fraction — $\sqrt{2}$, $\sqrt{3}$, $\sqrt[3]{5}$.
• Surds are named directly in the standard's list of methods, and they turn up in this paper for a specific reason: the quadratic formula produces them, and so does every modulus you calculate.
• An answer left as $\sqrt{20}$ instead of $2\sqrt{5}$, or with a surd still in the denominator, is not in simplest form and can lose the mark.

Simplifying a surd

To simplify $\sqrt{n}$:

• Find the largest perfect square that divides $n$.

• Split the root into two: $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$.

• Take the square root of the perfect square outside.

• Worked through:

  • $$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$$
  • $$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$$
  • $$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$$

• Note the rule works for division too: $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$.

The rule that does not exist

• $\sqrt{a + b}$ is not $\sqrt{a} + \sqrt{b}$.
• Check it once and never forget it: $\sqrt{9 + 16} = \sqrt{25} = 5$, but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$.
• Roots distribute over multiplication and division only, never over addition or subtraction.

Adding and subtracting surds

• Surds add like algebraic terms: you can only combine like surds, ones with the same number under the root.

• $3\sqrt{5} + 4\sqrt{5} = 7\sqrt{5}$, in exactly the way $3x + 4x = 7x$.

• $3\sqrt{5} + 4\sqrt{2}$ cannot be simplified — the surds are unlike.

• Always simplify first, because unlike surds often become like:

  • $$\sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$$

Multiplying surds

• Multiply the numbers outside, multiply the numbers inside, then simplify.

$$a\sqrt{b} \times c\sqrt{d} = ac\sqrt{bd}$$

• $$2\sqrt{3} \times 5\sqrt{6} = 10\sqrt{18} = 10 \times 3\sqrt{2} = 30\sqrt{2}$$

• Squaring a surd removes the root: $(\sqrt{7})^2 = 7$, and $(3\sqrt{2})^2 = 9 \times 2 = 18$.

• Expanding brackets works exactly as it does in algebra:

  • $$(2 + \sqrt{3})(5 - \sqrt{3}) = 10 - 2\sqrt{3} + 5\sqrt{3} - 3 = 7 + 3\sqrt{3}$$

Rationalising the denominator

• Rationalising means rewriting a fraction so no surd is left in the denominator.
• For a single surd, multiply top and bottom by that surd:

$$\frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$$

• For a two-term denominator, multiply top and bottom by the conjugate — the same expression with the middle sign flipped.

• The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$, and multiplying them gives a difference of two squares:

$$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$$

• The surd disappears because it gets squared. That is the entire trick.

• Worked through:

  • $$\dfrac{4}{3 + \sqrt{5}} = \dfrac{4}{3 + \sqrt{5}} \times \dfrac{3 - \sqrt{5}}{3 - \sqrt{5}} = \dfrac{4(3 - \sqrt{5})}{9 - 5} = \dfrac{4(3 - \sqrt{5})}{4} = 3 - \sqrt{5}$$

Why this matters for complex numbers

• The complex conjugate works by exactly the same mechanism. To divide by $3 + 2i$ you multiply by $3 - 2i$, and the bottom becomes $9 - 4i^2 = 9 + 4 = 13$.
• If you can rationalise a surd denominator fluently, division of complex numbers is the same move with $i^2 = -1$ instead of $(\sqrt{b})^2 = b$.
• The 2025 assessor report names both skills — rationalising denominators, and multiplying by a conjugate fraction — as fundamental, and lists failing to multiply surds correctly among the things Not Achieved candidates did.