Surds and Rationalisation

$\sqrt{9}$ is 3, so it is not a surd — the root came out exactly. $\sqrt{10}$ does not, and no decimal will ever capture it perfectly, so it stays as $\sqrt{10}$. That is the whole distinction: a $\text{surd}$ is a root whose value is irrational.

The two rules you need follow from indices, since $\sqrt{a}$ is just $a^{1/2}$. Roots pass through multiplication and division unchanged, which is what makes simplifying possible.

To simplify a surd, write the number as (largest square factor) × (rest), then split the root. $\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$. Using a smaller square factor is not wrong, only unfinished — $\sqrt{50}=\sqrt{2}\sqrt{25}$ reached the same place, but $\sqrt{72}=\sqrt{4}\sqrt{18}=2\sqrt{18}$ still has work left in it.

Once simplified, surds add and subtract exactly like algebraic terms: $3\sqrt{5}$ and $2\sqrt{5}$ combine to $5\sqrt{5}$ for the same reason $3x$ and $2x$ give $5x$. Unlike surds simply sit side by side.

Multiplication is straightforward: multiply the rational parts together and the surd parts together, then simplify. $2\sqrt{3}\times 5\sqrt{6}=10\sqrt{18}=10\times 3\sqrt{2}=30\sqrt{2}$.

Brackets expand exactly as they would in algebra, with $\sqrt{a}\times \sqrt{a}=a$ doing most of the tidying. That last identity is the engine of the next section, so it is worth saying out loud: a surd multiplied by itself stops being a surd.

Convention is that a final answer carries no surd on the bottom. For a single-surd denominator, multiply top and bottom by that surd — the bottom becomes rational and the value is unchanged, because you multiplied by 1 in disguise.

For a two-term denominator like $3-\sqrt{2}$, one surd is not enough; you need the $\text{conjugate}$, the same expression with the sign between the terms reversed. Multiplying by it triggers the difference of two squares from the previous section and clears the root in one move.