# 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 $\sqrt{{a}^{2}+8a+16}$ .

$a+4$

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