# 0.1 Surds

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## Surd calculations

There are several laws that make working with surds (or roots) easier. We will list them all and then explain where each rule comes from in detail.

$\begin{array}{ccc}\hfill \sqrt[n]{a}\sqrt[n]{b}& =& \sqrt[n]{ab}\hfill \\ \hfill \sqrt[n]{\frac{a}{b}}& =& \frac{\sqrt[n]{a}}{\sqrt[n]{b}}\hfill \\ \hfill \sqrt[n]{{a}^{m}}& =& {a}^{\frac{m}{n}}\hfill \end{array}$

## Surd law 1: $\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}$

It is often useful to look at a surd in exponential notation as it allows us to use the exponential laws we learnt in Grade 10. In exponential notation, $\sqrt[n]{a}={a}^{\frac{1}{n}}$ and $\sqrt[n]{b}={b}^{\frac{1}{n}}$ . Then,

$\begin{array}{ccc}\hfill \sqrt[n]{a}\sqrt[n]{b}& =& {a}^{\frac{1}{n}}{b}^{\frac{1}{n}}\hfill \\ \hfill & =& {\left(ab\right)}^{\frac{1}{n}}\hfill \\ \hfill & =& \sqrt[n]{ab}\hfill \end{array}$

Some examples using this law:

1. $\sqrt{16}×\sqrt{4}=\sqrt{64}=4$
2. $\sqrt{2}×\sqrt{32}=\sqrt{64}=8$
3. $\sqrt{{a}^{2}{b}^{3}}×\sqrt{{b}^{5}{c}^{4}}=\sqrt{{a}^{2}{b}^{8}{c}^{4}}=a{b}^{4}{c}^{2}$

## Surd law 2: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

If we look at $\sqrt[n]{\frac{a}{b}}$ in exponential notation and apply the exponential laws then,

$\begin{array}{ccc}\hfill \sqrt[n]{\frac{a}{b}}& =& {\left(\frac{a}{b}\right)}^{\frac{1}{n}}\hfill \\ \hfill & =& \frac{{a}^{\frac{1}{n}}}{{b}^{\frac{1}{n}}}\hfill \\ \hfill & =& \frac{\sqrt[n]{a}}{\sqrt[n]{b}}\hfill \end{array}$

Some examples using this law:

1. $\sqrt{12}÷\sqrt{3}=\sqrt{4}=2$
2. $\sqrt{24}÷\sqrt{3}=\sqrt{8}=2$
3. $\sqrt{{a}^{2}{b}^{13}}÷\sqrt{{b}^{5}}=\sqrt{{a}^{2}{b}^{8}}=a{b}^{4}$

## Surd law 3: $\sqrt[n]{{a}^{m}}={a}^{\frac{m}{n}}$

If we look at $\sqrt[n]{{a}^{m}}$ in exponential notation and apply the exponential laws then,

$\begin{array}{ccc}\hfill \sqrt[n]{{a}^{m}}& =& {\left({a}^{m}\right)}^{\frac{1}{n}}\hfill \\ \hfill & =& {a}^{\frac{m}{n}}\hfill \end{array}$

For example,

$\begin{array}{ccc}\hfill \sqrt{{2}^{3}}& =& {2}^{\frac{3}{6}}\hfill \\ & =& {2}^{\frac{1}{2}}\hfill \\ & =& \sqrt{2}\hfill \end{array}$

## Like and unlike surds

Two surds $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are called like surds if $m=n$ , otherwise they are called unlike surds . For example $\sqrt{2}$ and $\sqrt{3}$ are like surds, however $\sqrt{2}$ and $\sqrt{2}$ are unlike surds. An important thing to realise about the surd laws we have just learnt is that the surds in the laws are all like surds.

If we wish to use the surd laws on unlike surds, then we must first convert them into like surds. In order to do this we use the formula

$\sqrt[n]{{a}^{m}}=\sqrt[bn]{{a}^{bm}}$

to rewrite the unlike surds so that $bn$ is the same for all the surds.

Simplify to like surds as far as possible, showing all steps: $\sqrt{3}×\sqrt{5}$

1. $\begin{array}{ccc}& =& \sqrt{{3}^{5}}×\sqrt{{5}^{3}}\hfill \end{array}$
2. $\begin{array}{ccc}& =& \sqrt{{3}^{5}.{5}^{3}}\hfill \\ & =& \sqrt{243×125}\hfill \\ & =& \sqrt{30375}\hfill \end{array}$

## Simplest surd form

In most cases, when working with surds, answers are given in simplest surd form. For example,

$\begin{array}{ccc}\hfill \sqrt{50}& =& \sqrt{25×2}\hfill \\ & =& \sqrt{25}×\sqrt{2}\hfill \\ & =& 5\sqrt{2}\hfill \end{array}$

$5\sqrt{2}$ is the simplest surd form of $\sqrt{50}$ .

Rewrite $\sqrt{18}$ in the simplest surd form:

1. $\begin{array}{ccc}\hfill \sqrt{18}& =& \sqrt{2×9}\hfill \\ & =& \sqrt{2×3×3}\hfill \\ & =& \sqrt{2}×\sqrt{3×3}\hfill \\ & =& \sqrt{2}×\sqrt{{3}^{2}}\hfill \\ & =& 3\sqrt{2}\hfill \end{array}$

Simplify: $\sqrt{147}+\sqrt{108}$

1. $\begin{array}{ccc}\hfill \sqrt{147}+\sqrt{108}& =& \sqrt{49×3}+\sqrt{36×3}\hfill \\ & =& \sqrt{{7}^{2}×3}+\sqrt{{6}^{2}×3}\hfill \end{array}$
2. $\begin{array}{ccc}& =& 7\sqrt{3}+6\sqrt{3}\hfill \end{array}$
3. $\begin{array}{ccc}& =& 13\sqrt{3}\hfill \end{array}$

This video gives some examples of simplifying surds.

## Rationalising denominators

It is useful to work with fractions, which have rational denominators instead of surd denominators. It is possible to rewrite any fraction, which has a surd in the denominator as a fraction which has a rational denominator. We will now see how this can be achieved.

Any expression of the form $\sqrt{a}+\sqrt{b}$ (where $a$ and $b$ are rational) can be changed into a rational number by multiplying by $\sqrt{a}-\sqrt{b}$ (similarly $\sqrt{a}-\sqrt{b}$ can be rationalised by multiplying by $\sqrt{a}+\sqrt{b}$ ). This is because

$\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b$

which is rational (since $a$ and $b$ are rational).

If we have a fraction which has a denominator which looks like $\sqrt{a}+\sqrt{b}$ , then we can simply multiply both top and bottom by $\sqrt{a}-\sqrt{b}$ achieving a rational denominator.

$\begin{array}{ccc}\hfill \frac{c}{\sqrt{a}+\sqrt{b}}& =& \frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}×\frac{c}{\sqrt{a}+\sqrt{b}}\hfill \\ \hfill & =& \frac{c\sqrt{a}-c\sqrt{b}}{a-b}\hfill \end{array}$

or similarly

$\begin{array}{ccc}\hfill \frac{c}{\sqrt{a}-\sqrt{b}}& =& \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}×\frac{c}{\sqrt{a}-\sqrt{b}}\hfill \\ \hfill & =& \frac{c\sqrt{a}+c\sqrt{b}}{a-b}\hfill \end{array}$

Rationalise the denominator of: $\frac{5x-16}{\sqrt{x}}$

1. To get rid of $\sqrt{x}$ in the denominator, you can multiply it out by another $\sqrt{x}$ . This "rationalises" the surd in the denominator. Note that $\frac{\sqrt{x}}{\sqrt{x}}$ = 1, thus the equation becomes rationalised by multiplying by 1 and thus still says the same thing.

$\begin{array}{c}\hfill \frac{5x-16}{\sqrt{x}}×\frac{\sqrt{x}}{\sqrt{x}}\end{array}$
2. The surd is expressed in the numerator which is the prefered way to write expressions. (That's why denominators get rationalised.)

$\begin{array}{ccc}& & \frac{5x\sqrt{x}-16\sqrt{x}}{x}\hfill \\ & =& \frac{\left(\sqrt{x}\right)\left(5x-16\right)}{x}\hfill \end{array}$

Rationalise the following: $\frac{5x-16}{\sqrt{y}-10}$

1. $\begin{array}{c}\hfill \frac{5x-16}{\sqrt{y}-10}×\frac{\sqrt{y}+10}{\sqrt{y}+10}\end{array}$
2. $\begin{array}{c}\hfill \frac{5x\sqrt{y}-16\sqrt{y}+50x-160}{y-100}\end{array}$
3. All the terms in the numerator are different and cannot be simplified and the denominator does not have any surds in it anymore.

Simplify the following: $\frac{y-25}{\sqrt{y}+5}$

1. $\frac{y-25}{\sqrt{y}+5}×\frac{\sqrt{y}-5}{\sqrt{y}-5}$
2. $\begin{array}{ccc}\hfill \frac{y\sqrt{y}-25\sqrt{y}-5y+125}{y-25}& =& \frac{\sqrt{y}\left(y-25\right)-5\left(y-25\right)}{\left(y-25\right)}\hfill \\ & =& \frac{\left(y-25\right)\left(\sqrt{y}-25\right)}{\left(y-25\right)}\hfill \\ & =& \sqrt{y}-25\hfill \end{array}$

The following video explains some of the concepts of rationalising the denominator.

## End of chapter exercises

1. Expand:
$\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)$
2. Rationalise the denominator:
$\frac{10}{\sqrt{x}-\frac{1}{x}}$
3. Write as a single fraction:
$\frac{3}{2\sqrt{x}}+\sqrt{x}$
4. Write in simplest surd form:
1. $\sqrt{72}$
2. $\sqrt{45}+\sqrt{80}$
3. $\frac{\sqrt{48}}{\sqrt{12}}$
4. $\frac{\sqrt{18}÷\sqrt{72}}{\sqrt{8}}$
5. $\frac{4}{\left(\sqrt{8}÷\sqrt{2}\right)}$
6. $\frac{16}{\left(\sqrt{20}÷\sqrt{12}\right)}$
5. Expand and simplify:
${\left(2+\sqrt{2}\right)}^{2}$
6. Expand and simplify:
$\left(2+\sqrt{2}\right)\left(1+\sqrt{8}\right)$
7. Expand and simplify:
$\left(1+\sqrt{3}\right)\left(1+\sqrt{8}+\sqrt{3}\right)$
8. Rationalise the denominator:
$\frac{y-4}{\sqrt{y}-2}$
9. Rationalise the denominator:
$\frac{2x-20}{\sqrt{y}-\sqrt{10}}$
10. Prove (without the use of a calculator) that:
$\sqrt{\frac{8}{3}}+5\sqrt{\frac{5}{3}}-\sqrt{\frac{1}{6}}=\frac{13}{2}\sqrt{\frac{2}{3}}$
11. Simplify, without use of a calculator:
$\frac{\sqrt{98}-\sqrt{8}}{\sqrt{50}}$
12. Simplify, without use of a calculator:
$\sqrt{5}\left(\sqrt{45}+2\sqrt{80}\right)$
13. Write the following with a rational denominator:
$\frac{\sqrt{5}+2}{\sqrt{5}}$
14. Simplify:
$\sqrt{98{x}^{6}}+\sqrt{128{x}^{6}}$
15. Evaluate without using a calculator: ${\left(2,-,\frac{\sqrt{7}}{2}\right)}^{\frac{1}{2}}\phantom{\rule{0.222222em}{0ex}}.\phantom{\rule{0.222222em}{0ex}}{\left(2,+,\frac{\sqrt{7}}{2}\right)}^{\frac{1}{2}}$
16. The use of a calculator is not permissible in this question. Simplify completely by showing all your steps: ${3}^{-\frac{1}{2}}\left[\sqrt{12},+,\sqrt{\left(3\sqrt{3}\right)}\right]$
17. Fill in the blank surd-form number which will make the following equation a true statement: $-3\sqrt{6}×-2\sqrt{24}=-\sqrt{18}×\mathrm{______}$

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