Surds

What a surd is Extended

Some square roots are whole numbers or fractions: $\sqrt{49}=7$ and $\sqrt{\frac{9}{4}}=\frac{3}{2}$. Others are not. $\sqrt{2}$ is $1.4142135\dots$, and the decimals never end and never repeat, because $\sqrt{2}$ is irrational. A root like this, left written as a root, is called a surd.

So $\sqrt{2}$, $\sqrt{5}$ and $3\sqrt{7}$ are surds, but $\sqrt{49}$ is not, because it is exactly 7.

Why keep the root sign instead of using a decimal? Because the surd is exact. $1.414$ is only an approximation to $\sqrt{2}$, and if you square it you get $1.999396$, not 2. When a question asks for an exact answer, or the answer in surd form, a decimal scores nothing. On the non-calculator paper, surds are also simply the only way to write these numbers.

The rules for roots Extended

Two rules do almost all the work. For positive numbers $a$ and $b$:

• $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$, for example $\sqrt{6}\times \sqrt{5}=\sqrt{30}$;
• $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$, for example $\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}$.

A special case of the first rule is $\sqrt{a}\times \sqrt{a}=a$. So $\sqrt{7}\times \sqrt{7}=7$ and $(\sqrt{7}{)}^2=7$.

Why is the first rule true? Square the right-hand side: $(\sqrt{a}\times \sqrt{b}{)}^2=\sqrt{a}\times \sqrt{a}\times \sqrt{b}\times \sqrt{b}=ab$. So $\sqrt{a}\times \sqrt{b}$ is the positive number whose square is $ab$, which is exactly what $\sqrt{ab}$ means.

There is no rule like this for adding. $\sqrt{9+16}=\sqrt{25}=5$, but $\sqrt{9}+\sqrt{16}=3+4=7$. So $\sqrt{a+b}$ is not $\sqrt{a}+\sqrt{b}$.

Simplifying a surd Extended

A surd is in its simplest form when the number under the root has no square factor left (no factor of 4, 9, 16, 25, 36, 49, 64, 81, 100, ...).

Method: write the number as (largest square factor) $\times$ (what is left), then take the square root of the square factor outside.

The syllabus example: $\sqrt{20}=\sqrt{4\times 5}=\sqrt{4}\times \sqrt{5}=2\sqrt{5}$.

Look for the largest square factor. $\sqrt{72}=\sqrt{4\times 18}=2\sqrt{18}$ is true but not finished, because 18 still has the factor 9: $2\sqrt{18}=2\times 3\sqrt{2}=6\sqrt{2}$. Going straight to $\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$ is quicker. Working through the square numbers from the largest that could fit is a reliable way to find it.

Adding and subtracting surds Extended

Surds can be added or subtracted only when the root is the same, just as $5x+3x=8x$ but $5x+3y$ cannot be combined: $5\sqrt{2}+3\sqrt{2}=8\sqrt{2}$, but $\sqrt{2}+\sqrt{3}$ stays as it is.

Surds that look different often become like surds once they are simplified. Simplify each one first, then collect them.

The syllabus example shows this: $\sqrt{200}-\sqrt{32}=10\sqrt{2}-4\sqrt{2}=6\sqrt{2}$.

Multiplying surds and expanding brackets Extended

To multiply, multiply the whole numbers together and the roots together: $2\sqrt{3}\times 5\sqrt{6}=10\sqrt{18}=10\times 3\sqrt{2}=30\sqrt{2}$.

Expand brackets exactly as in algebra, multiplying every term in one bracket by every term in the other, and remember that $\sqrt{a}\times \sqrt{a}=a$. Then collect the whole numbers and the surds separately.

Two patterns are worth knowing:

• A perfect square: $(a+\sqrt{b}{)}^2=a^2+2a\sqrt{b}+b$. There are three terms, not two: $(2+\sqrt{3}{)}^2$ is not $4+3$.
• Conjugates: $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$. The surd terms cancel and the answer is a whole number. For example $(5+\sqrt{2})(5-\sqrt{2})=25-2=23$. This is the difference of two squares, and it is the key to rationalising.

Rationalising the denominator Extended

A fraction with a surd on the bottom, such as $\frac{10}{\sqrt{5}}$, is not in its final form. Rationalising the denominator means rewriting it with a whole number on the bottom. It does not change the value of the fraction, because you multiply the top and the bottom by the same thing, which is the same as multiplying by 1.

A single surd on the bottom. Multiply top and bottom by that surd, since $\sqrt{a}\times \sqrt{a}=a$. The syllabus example:

$$\frac{10}{\sqrt{5}}=\frac{10\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}=\frac{10\sqrt{5}}{5}=2\sqrt{5}$$

A bracket such as $a+\sqrt{b}$ on the bottom. Multiplying by $\sqrt{b}$ would not remove the root. Instead multiply top and bottom by the conjugate, the same bracket with the sign in front of the surd changed. The difference of two squares then clears the root: $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$.

The syllabus example is $\frac{1}{-1+\sqrt{3}}$. Multiply top and bottom by $1+\sqrt{3}$ (changing the sign of the $-1$ works just as well as changing the sign of the surd):

$$\frac{1\times (1+\sqrt{3})}{(-1+\sqrt{3})(1+\sqrt{3})}=\frac{1+\sqrt{3}}{-1-\sqrt{3}+\sqrt{3}+3}=\frac{1+\sqrt{3}}{2}$$

Always expand the denominator fully and check that the surd terms cancel. If they do not, you have used the wrong multiplier.