# 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: $-3\sqrt{6}×-2\sqrt{24}=-\sqrt{18}×\mathrm{______}$

where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
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da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
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Alexandre
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Rafiq
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Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
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Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
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Mahi
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Rafiq
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Anam
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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write examples of Nano molecule?
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brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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Kyle
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biomolecules are e building blocks of every organics and inorganic materials.
Joe
Other chapter Q/A we can ask

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