Laws of surds, simplifying, rationalising denominators with conjugates, equating rational and irrational parts, and square roots of expressions like 7 + 4√3.
A surd is a root of a rational number that is itself irrational: $\sqrt{2}$, $\sqrt{18}$, $\sqrt[3]{5}$. The number $\sqrt{9} = 3$ is not a surd, because it is rational. In this lesson $\sqrt{a}$ always means the non-negative square root, so $\sqrt{9} = 3$ (not $\pm 3$).
For $a \geq 0$, $b \geq 0$:
Common mistake: $\sqrt{a + b}$ is not $\sqrt{a} + \sqrt{b}$. For example $\sqrt{9 + 16} = 5$ but $\sqrt{9} + \sqrt{16} = 7$.
Simplifying. Take out the largest square factor: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$, and $\sqrt{50} = 5\sqrt{2}$. Like surds have the same surd part and can be added: $6\sqrt{2} - 5\sqrt{2} + \sqrt{2} = 2\sqrt{2}$, but $\sqrt{2} + \sqrt{3}$ cannot be combined.
We prefer to have no surd in a denominator, which makes numbers easier to compare and add.
Type 1: a single surd. $\dfrac{6}{\sqrt{3}} = \dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$.
Type 2: two terms. Use the conjugate: the conjugate of $p + q\sqrt{n}$ is $p - q\sqrt{n}$ (change one sign). Their product has no surd, by the difference of two squares: $$(p + q\sqrt{n})(p - q\sqrt{n}) = p^2 - q^2 n.$$ So $\dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \dfrac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$.
Example: Write $\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}}$ in the form $p + q\sqrt{2}$ with $p, q$ rational. Multiply top and bottom by the conjugate $3 + \sqrt{2}$: $\dfrac{(3 + \sqrt{2})^2}{(3 - \sqrt{2})(3 + \sqrt{2})} = \dfrac{9 + 6\sqrt{2} + 2}{9 - 2} = \dfrac{11 + 6\sqrt{2}}{7} = \dfrac{11}{7} + \dfrac{6}{7}\sqrt{2}$.
Example: Simplify $\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + 2}$. Rationalise each term. Each denominator times its conjugate gives $1$: $\dfrac{1}{1 + \sqrt{2}} = \dfrac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1$, $\ \dfrac{1}{\sqrt{2} + \sqrt{3}} = \dfrac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$, $\ \dfrac{1}{\sqrt{3} + 2} = \dfrac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$. The sum is $(\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3}) = 1$, because the middle terms cancel in pairs.
Result. If $a, b, c, d$ are rational and $\sqrt{n}$ is irrational, and $a + b\sqrt{n} = c + d\sqrt{n}$, then $a = c$ and $b = d$.
Why. Rearrange to $(b - d)\sqrt{n} = c - a$. If $b \neq d$ then $\sqrt{n} = \dfrac{c - a}{b - d}$ would be rational, which is false. So $b = d$, and then $a = c$. $\square$
Example: Find rational numbers $a, b$ such that $\dfrac{5 + \sqrt{3}}{2 - \sqrt{3}} = a + b\sqrt{3}$. Rationalise: $\dfrac{(5 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \dfrac{10 + 5\sqrt{3} + 2\sqrt{3} + 3}{4 - 3} = 13 + 7\sqrt{3}$. Equating the rational and the irrational parts gives $a = 13$ and $b = 7$.
To find $\sqrt{p + q\sqrt{n}}$, try $\sqrt{x} + \sqrt{y}$ with $x, y > 0$. Squaring, $(\sqrt{x} + \sqrt{y})^2 = (x + y) + 2\sqrt{xy}$. So we need $x + y = p$ and $2\sqrt{xy} = q\sqrt{n}$, that is $4xy = q^2 n$.
Example: Find $\sqrt{7 + 4\sqrt{3}}$. Write $4\sqrt{3} = 2\sqrt{12}$, so we need $x + y = 7$ and $xy = 12$. The numbers $4$ and $3$ work. Check: $(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$. Since $2 + \sqrt{3} > 0$, $\sqrt{7 + 4\sqrt{3}} = 2 + \sqrt{3}$.
For a difference such as $\sqrt{7 - 4\sqrt{3}}$ the answer is $2 - \sqrt{3}$, which is positive: take the larger root first so that the result is not negative.
Exam tip: Whenever you see $a \pm \sqrt{b}$ in a denominator, multiply by the conjugate straight away and expand the numerator carefully. Part (b) of such a question often asks you to "hence" evaluate, or to find $a$ and $b$: finish by equating the rational and irrational parts.