# 9.4 Division of square root expressions

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The distinction between the principal square root of the number x and the secondary square root of the number x is made by explanation and by example. The simplification of the radical expressions that both involve and do not involve fractions is shown in many detailed examples; this is followed by an explanation of how and why radicals are eliminated from the denominator of a radical expression. Real-life applications of radical equations have been included, such as problems involving daily output, daily sales, electronic resonance frequency, and kinetic energy.Objectives of this module: be able to use the division property of square roots, the method of rationalizing the denominator, and conjugates to divide square roots.

## Overview

• The Division Property of Square Roots
• Rationalizing the Denominator
• Conjugates and Rationalizing the Denominator

## The division property of square roots

In our work with simplifying square root expressions, we noted that

$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$

Since this is an equation, we may write it as

$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$

To divide two square root expressions, we use the division property of square roots.

## The division property $\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$

$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$

The quotient of the square roots is the square root of the quotient.

## Rationalizing the denominator

As we can see by observing the right side of the equation governing the division of square roots, the process may produce a fraction in the radicand. This means, of course, that the square root expression is not in simplified form. It is sometimes more useful to rationalize the denominator of a square root expression before actually performing the division.

## Sample set a

Simplify the square root expressions.

$\sqrt{\frac{3}{7}}.$

This radical expression is not in simplified form since there is a fraction under the radical sign. We can eliminate this problem using the division property of square roots.

$\sqrt{\frac{3}{7}}=\frac{\sqrt{3}}{\sqrt{7}}=\frac{\sqrt{3}}{\sqrt{7}}·\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{3}\sqrt{7}}{7}=\frac{\sqrt{21}}{7}$

$\frac{\sqrt{5}}{\sqrt{3}}.$

A direct application of the rule produces $\sqrt{\frac{5}{3}},$ which must be simplified. Let us rationalize the denominator before we perform the division.

$\frac{\sqrt{5}}{\sqrt{3}}=\frac{\sqrt{5}}{\sqrt{3}}·\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{5}\sqrt{3}}{3}=\frac{\sqrt{15}}{3}$

$\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{\frac{21}{7}}=\sqrt{3}.$

The rule produces the quotient quickly. We could also rationalize the denominator first and produce the same result.

$\frac{\sqrt{21}}{\sqrt{7}}=\frac{\sqrt{21}}{7}·\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{21·7}}{7}=\frac{\sqrt{3·7·7}}{7}=\frac{\sqrt{3·{7}^{2}}}{7}=\frac{7\sqrt{3}}{7}=\sqrt{3}$

$\frac{\sqrt{80{x}^{9}}}{\sqrt{5{x}^{4}}}=\sqrt{\frac{80{x}^{9}}{5{x}^{4}}}=\sqrt{16{x}^{5}}=\sqrt{16}\sqrt{{x}^{4}x}=4{x}^{2}\sqrt{x}$

$\frac{\sqrt{50{a}^{3}{b}^{7}}}{\sqrt{5a{b}^{5}}}=\sqrt{\frac{50{a}^{3}{b}^{7}}{5a{b}^{5}}}=\sqrt{10{a}^{2}{b}^{2}}=ab\sqrt{10}$

$\frac{\sqrt{5a}}{\sqrt{b}}.$

Some observation shows that a direct division of the radicands will produce a fraction. This suggests that we rationalize the denominator first.

$\frac{\sqrt{5a}}{\sqrt{b}}=\frac{\sqrt{5a}}{\sqrt{b}}·\frac{\sqrt{b}}{\sqrt{b}}=\frac{\sqrt{5a}\sqrt{b}}{b}=\frac{\sqrt{5ab}}{b}$

$\frac{\sqrt{m-6}}{\sqrt{m+2}}=\frac{\sqrt{m-6}}{\sqrt{m+2}}·\frac{\sqrt{m+2}}{\sqrt{m+2}}=\frac{\sqrt{{m}^{2}-4m-12}}{m+2}$

$\frac{\sqrt{{y}^{2}-y-12}}{\sqrt{y+3}}=\sqrt{\frac{{y}^{2}-y-12}{y+3}}=\sqrt{\frac{\left(y+3\right)\left(y-4\right)}{\left(y+3\right)}}=\sqrt{\frac{\overline{)\left(y+3\right)}\left(y-4\right)}{\overline{)\left(y+3\right)}}}=\sqrt{y-4}$

## Practice set a

Simplify the square root expressions.

$\frac{\sqrt{26}}{\sqrt{13}}$

$\sqrt{2}$

$\frac{\sqrt{7}}{\sqrt{3}}$

$\frac{\sqrt{21}}{3}$

$\frac{\sqrt{80{m}^{5}{n}^{8}}}{\sqrt{5{m}^{2}n}}$

$4m{n}^{3}\sqrt{mn}$

$\frac{\sqrt{196{\left(x+7\right)}^{8}}}{\sqrt{2{\left(x+7\right)}^{3}}}$

$7{\left(x+7\right)}^{2}\sqrt{2\left(x+7\right)}$

$\frac{\sqrt{n+4}}{\sqrt{n-5}}$

$\frac{\sqrt{{n}^{2}-n-20}}{n-5}$

$\frac{\sqrt{{a}^{2}-6a+8}}{\sqrt{a-2}}$

$\sqrt{a-4}$

$\frac{\sqrt{{x}^{3}{}^{n}}}{\sqrt{{x}^{n}}}$

${x}^{n}$

$\frac{\sqrt{{a}^{3m-5}}}{\sqrt{{a}^{m-1}}}$

${a}^{m-2}$

## Conjugates and rationalizing the denominator

To perform a division that contains a binomial in the denominator, such as $\frac{3}{4+\sqrt{6}},$ we multiply the numerator and denominator by a conjugate of the denominator.

## Conjugate

A conjugate of the binomial $a+b$ is $a-b$ . Similarly, a conjugate of $a-b$ is $a+b$ .

Notice that when the conjugates $a+b$ and $a-b$ are multiplied together, they produce a difference of two squares.

$\left(a+b\right)\left(a-b\right)={a}^{2}-ab+ab-{b}^{2}={a}^{2}-{b}^{2}$

This principle helps us eliminate square root radicals, as shown in these examples that illustrate finding the product of conjugates.

$\begin{array}{lll}\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)\hfill & =\hfill & {5}^{2}-{\left(\sqrt{2}\right)}^{2}\hfill \\ \hfill & =\hfill & 25-2\hfill \\ \hfill & =\hfill & 23\hfill \end{array}$

$\begin{array}{lll}\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{6}+\sqrt{7}\right)\hfill & =\hfill & {\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}\hfill \\ \hfill & =\hfill & 6-7\hfill \\ \hfill & =\hfill & -1\hfill \end{array}$

## Sample set b

Simplify the following expressions.

$\frac{3}{4+\sqrt{6}}$ .

The conjugate of the denominator is $4-\sqrt{6.}$ Multiply the fraction by 1 in the form of $\frac{4-\sqrt{6}}{4-\sqrt{6}}$ . $\begin{array}{lll}\frac{3}{4+\sqrt{6}}·\frac{4-\sqrt{6}}{4-\sqrt{6}}\hfill & =\hfill & \frac{3\left(4-\sqrt{6}\right)}{{4}^{2}-{\left(\sqrt{6}\right)}^{2}}\hfill \\ \hfill & =\hfill & \frac{12-3\sqrt{6}}{16-6}\hfill \\ \hfill & =\hfill & \frac{12-3\sqrt{6}}{10}\hfill \end{array}$

$\frac{\sqrt{2x}}{\sqrt{3}-\sqrt{5x}}.$

The conjugate of the denominator is $\sqrt{3}+\sqrt{5x.}$ Multiply the fraction by 1 in the form of $\frac{\sqrt{3}+\sqrt{5x}}{\sqrt{3}+\sqrt{5x}}.$

$\begin{array}{lll}\frac{\sqrt{2x}}{\sqrt{3}-\sqrt{5x}}·\frac{\sqrt{3}+\sqrt{5x}}{\sqrt{3}+\sqrt{5x}}\hfill & =\hfill & \frac{\sqrt{2x}\left(\sqrt{3}+\sqrt{5x}\right)}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{5x}\right)}^{2}}\hfill \\ \hfill & =\hfill & \frac{\sqrt{2x}\sqrt{3}+\sqrt{2x}\sqrt{5x}}{3-5x}\hfill \\ \hfill & =\hfill & \frac{\sqrt{6x}+\sqrt{10{x}^{2}}}{3-5x}\hfill \\ \hfill & =\hfill & \frac{\sqrt{6x}+x\sqrt{10}}{3-5x}\hfill \end{array}$

## Practice set b

Simplify the following expressions.

$\frac{5}{9+\sqrt{7}}$

$\frac{45-5\sqrt{7}}{74}$

$\frac{-2}{1-\sqrt{3x}}$

$\frac{-2-2\sqrt{3x}}{1-3x}$

$\frac{\sqrt{8}}{\sqrt{3x}+\sqrt{2x}}$

$\frac{2\sqrt{6x}-4\sqrt{x}}{x}$

$\frac{\sqrt{2m}}{m-\sqrt{3m}}$

$\frac{\sqrt{2m}+\sqrt{6}}{m-3}$

## Exercises

For the following problems, simplify each expressions.

$\frac{\sqrt{28}}{\sqrt{2}}$

$\sqrt{14}$

$\frac{\sqrt{200}}{\sqrt{10}}$

$\frac{\sqrt{28}}{\sqrt{7}}$

2

$\frac{\sqrt{96}}{\sqrt{24}}$

$\frac{\sqrt{180}}{\sqrt{5}}$

6

$\frac{\sqrt{336}}{\sqrt{21}}$

$\frac{\sqrt{162}}{\sqrt{18}}$

3

$\sqrt{\frac{25}{9}}$

$\sqrt{\frac{36}{35}}$

$\frac{6\sqrt{35}}{35}$

$\sqrt{\frac{225}{16}}$

$\sqrt{\frac{49}{225}}$

$\frac{7}{15}$

$\sqrt{\frac{3}{5}}$

$\sqrt{\frac{3}{7}}$

$\frac{\sqrt{21}}{7}$

$\sqrt{\frac{1}{2}}$

$\sqrt{\frac{5}{2}}$

$\frac{\sqrt{10}}{2}$

$\sqrt{\frac{11}{25}}$

$\sqrt{\frac{15}{36}}$

$\frac{\sqrt{15}}{6}$

$\sqrt{\frac{5}{16}}$

$\sqrt{\frac{7}{25}}$

$\frac{\sqrt{7}}{5}$

$\sqrt{\frac{32}{49}}$

$\sqrt{\frac{50}{81}}$

$\frac{5\sqrt{2}}{9}$

$\frac{\sqrt{125{x}^{5}}}{\sqrt{5{x}^{3}}}$

$\frac{\sqrt{72{m}^{7}}}{\sqrt{2{m}^{3}}}$

$6{m}^{2}$

$\frac{\sqrt{162{a}^{11}}}{\sqrt{2{a}^{5}}}$

$\frac{\sqrt{75{y}^{10}}}{\sqrt{3{y}^{4}}}$

$5{y}^{3}$

$\frac{\sqrt{48{x}^{9}}}{\sqrt{3{x}^{2}}}$

$\frac{\sqrt{125{a}^{14}}}{\sqrt{5{a}^{5}}}$

$5{a}^{4}\sqrt{a}$

$\frac{\sqrt{27{a}^{10}}}{\sqrt{3{a}^{5}}}$

$\frac{\sqrt{108{x}^{21}}}{\sqrt{3{x}^{4}}}$

$6{x}^{8}\sqrt{x}$

$\frac{\sqrt{48{x}^{6}{y}^{7}}}{\sqrt{3xy}}$

$\frac{\sqrt{45{a}^{3}{b}^{8}{c}^{2}}}{\sqrt{5a{b}^{2}c}}$

$3a{b}^{3}\sqrt{c}$

$\frac{\sqrt{66{m}^{12}{n}^{15}}}{\sqrt{11m{n}^{8}}}$

$\frac{\sqrt{30{p}^{5}{q}^{14}}}{\sqrt{5{q}^{7}}}$

${p}^{2}{q}^{3}\sqrt{6pq}$

$\frac{\sqrt{b}}{\sqrt{5}}$

$\frac{\sqrt{5x}}{\sqrt{2}}$

$\frac{\sqrt{10x}}{2}$

$\frac{\sqrt{2{a}^{3}b}}{\sqrt{14a}}$

$\frac{\sqrt{3{m}^{4}{n}^{3}}}{\sqrt{6m{n}^{5}}}$

$\frac{m\sqrt{2m}}{2n}$

$\frac{\sqrt{5{\left(p-q\right)}^{6}{\left(r+s\right)}^{4}}}{\sqrt{25{\left(r+s\right)}^{3}}}$

$\frac{\sqrt{m\left(m-6\right)-{m}^{2}+6m}}{\sqrt{3m-7}}$

0

$\frac{\sqrt{r+1}}{\sqrt{r-1}}$

$\frac{\sqrt{s+3}}{\sqrt{s-3}}$

$\frac{\sqrt{{s}^{2}-9}}{s-3}$

$\frac{\sqrt{{a}^{2}+3a+2}}{\sqrt{a+1}}$

$\frac{\sqrt{{x}^{2}-10x+24}}{\sqrt{x-4}}$

$\sqrt{x-6}$

$\frac{\sqrt{{x}^{2}-2x-8}}{\sqrt{x+2}}$

$\frac{\sqrt{{x}^{2}-4x+3}}{\sqrt{x-3}}$

$\sqrt{x-1}$

$\frac{\sqrt{2{x}^{2}-x-1}}{\sqrt{x-1}}$

$\frac{-5}{4+\sqrt{5}}$

$\frac{-20+5\sqrt{5}}{11}$

$\frac{1}{1+\sqrt{x}}$

$\frac{2}{1-\sqrt{a}}$

$\frac{2\left(1+\sqrt{a}\right)}{1-a}$

$\frac{-6}{\sqrt{5}-1}$

$\frac{-6}{\sqrt{7}+2}$

$-2\left(\sqrt{7}-2\right)$

$\frac{3}{\sqrt{3}-\sqrt{2}}$

$\frac{4}{\sqrt{6}+\sqrt{2}}$

$\sqrt{6}-\sqrt{2}$

$\frac{\sqrt{5}}{\sqrt{8}-\sqrt{6}}$

$\frac{\sqrt{12}}{\sqrt{12}-\sqrt{8}}$

$3+\sqrt{6}$

$\frac{\sqrt{7x}}{2-\sqrt{5x}}$

$\frac{\sqrt{6y}}{1+\sqrt{3y}}$

$\frac{\sqrt{6y}-3y\sqrt{2}}{1-3y}$

$\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}$

$\frac{a-\sqrt{ab}}{a-b}$

$\frac{\sqrt{{8}^{3}{b}^{5}}}{4-\sqrt{2ab}}$

$\frac{\sqrt{7x}}{\sqrt{5x}+\sqrt{x}}$

$\frac{\sqrt{35}-\sqrt{7}}{4}$

$\frac{\sqrt{3y}}{\sqrt{2y}-\sqrt{y}}$

## Exercises for review

( [link] ) Simplify ${x}^{8}{y}^{7}\left(\frac{{x}^{4}{y}^{8}}{{x}^{3}{y}^{4}}\right).$

${x}^{9}{y}^{11}$

( [link] ) Solve the compound inequality $-8\le 7-5x\le -23.$

( [link] ) Construct the graph of $y=\frac{2}{3}x-4.$

( [link] ) The symbol $\sqrt{x}$ represents which square root of the number $x,\text{\hspace{0.17em}}x\ge 0$ ?

( [link] ) Simplify $\sqrt{{a}^{2}+8a+16}$ .

$a+4$

#### Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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