A surd is a root — like $\sqrt{2}$ or $\sqrt{5}$ — that cannot be simplified to a whole number or exact fraction, because the value underneath is irrational. Keeping an answer as a surd instead of rounding it to a decimal preserves perfect accuracy, which is exactly what "show that" questions demand.

Find the largest perfect-square factor of the number underneath, split it off using $\sqrt{a}\times\sqrt{b} = \sqrt{ab}$, and take its square root.

Only like surds (the same value under the root) can be combined, just like collecting like terms:

Remember that a surd never splits over addition: $\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$ — always simplify inside the root first, e.g. $\sqrt{16+9}=\sqrt{25}=5$, not $\sqrt{16}+\sqrt{9}=7$.

Worked Example

Simplify $\sqrt{45} + \sqrt{80} - \sqrt{20}$, giving your answer in the form $k\sqrt{5}$.

1. $\sqrt{45} = \sqrt{9\times5} = 3\sqrt{5}$
2. $\sqrt{80} = \sqrt{16\times5} = 4\sqrt{5}$
3. $\sqrt{20} = \sqrt{4\times5} = 2\sqrt{5}$
4. Combine: $3\sqrt{5}+4\sqrt{5}-2\sqrt{5} = 5\sqrt{5}$, so $k=5$.

Expand exactly as you would with algebraic brackets, multiplying each term in the first bracket by each term in the second, then simplify using $\sqrt{a}\times\sqrt{a}=a$.

Worked Example

Expand and simplify $(3+\sqrt{2})(2-\sqrt{2})$.

1. Multiply out term by term: $3\times2 + 3\times(-\sqrt{2}) + \sqrt{2}\times2 + \sqrt{2}\times(-\sqrt{2})$.
2. Simplify each term: $6 - 3\sqrt{2} + 2\sqrt{2} - 2$.
3. Combine the whole-number terms and the surd terms separately: $(6-2) + (-3\sqrt{2}+2\sqrt{2}) = 4-\sqrt{2}$.

Squaring a surd like $\sqrt{2}$ on a calculator gives a rounded decimal, which loses information the moment you write it down. Keeping expressions in surd form throughout a multi-step problem preserves total accuracy, which is exactly what a mark scheme checks for whenever it asks for an answer "in the form $a+b\sqrt{c}$" or "leave your answer as a surd".

Multiply top and bottom by that surd, so the denominator becomes rational:

Multiply top and bottom by the denominator's conjugate (same terms, opposite middle sign). This triggers the difference-of-two-squares identity, which removes the root from the denominator entirely.

Worked Example

Express $\dfrac{4}{2+\sqrt{2}}$ in the form $a+b\sqrt{2}$.

1. The conjugate of $2+\sqrt{2}$ is $2-\sqrt{2}$.
2. Multiply top and bottom by the conjugate: $\dfrac{4}{2+\sqrt{2}}\times\dfrac{2-\sqrt{2}}{2-\sqrt{2}} = \dfrac{4(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$.
3. Expand the numerator: $4(2-\sqrt{2}) = 8-4\sqrt{2}$.
4. Expand the denominator (difference of two squares): $2^2-(\sqrt{2})^2 = 4-2=2$.
5. Divide through by $2$: $\dfrac{8-4\sqrt{2}}{2} = 4-2\sqrt{2}$, so $a=4$, $b=-2$.