9.2 Simplifying 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 identify a perfect square, be familiar with the product and quotient properties of square roots, be able to simplify square roots involving and not involving fractions.

Overview

• Perfect Squares
• The Product Property of Square Roots
• The Quotient Property of Square Roots
• Square Roots Not Involving Fractions
• Square Roots Involving Fractions

To begin our study of the process of simplifying a square root expression, we must note three facts: one fact concerning perfect squares and two concerning properties of square roots.

Perfect squares

Real numbers that are squares of rational numbers are called perfect squares. The numbers 25 and $\frac{1}{4}$ are examples of perfect squares since $25={5}^{2}$ and $\frac{1}{4}={\left(\frac{1}{2}\right)}^{2},$ and 5 and $\frac{1}{2}$ are rational numbers. The number 2 is not a perfect square since $2={\left(\sqrt{2}\right)}^{2}$ and $\sqrt{2}$ is not a rational number.

Although we will not make a detailed study of irrational numbers, we will make the following observation:

Any indicated square root whose radicand is not a perfect square is an irrational number.

The numbers $\sqrt{6},\text{\hspace{0.17em}}\sqrt{15},$ and $\sqrt{\frac{3}{4}}$ are each irrational since each radicand $\left(6,\text{\hspace{0.17em}}15,\text{\hspace{0.17em}}\frac{3}{4}\right)$ is not a perfect square.

The product property of square roots

Notice that

$\sqrt{9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}4}=\sqrt{36}=6$      and
$\sqrt{9}\text{\hspace{0.17em}}\sqrt{4}=3\text{\hspace{0.17em}}·2\text{\hspace{0.17em}}=6$

Since both $\sqrt{9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}4}$ and $\sqrt{9}\text{\hspace{0.17em}}\sqrt{4}$ equal 6, it must be that

$\sqrt{9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}4}=\sqrt{9}\sqrt{4}$

The product property $\sqrt{xy}=\sqrt{x}\text{\hspace{0.17em}}\sqrt{y}$

This suggests that in general, if $x$ and $y$ are positive real numbers,

$\sqrt{xy}=\sqrt{x}\text{\hspace{0.17em}}\sqrt{y}$

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

The quotient property of square roots

We can suggest a similar rule for quotients. Notice that

$\sqrt{\frac{36}{4}}=\sqrt{9}=3$      and
$\frac{\sqrt{36}}{\sqrt{4}}=\frac{6}{2}=3$

Since both $\frac{36}{4}$ and $\frac{\sqrt{36}}{\sqrt{4}}$ equal 3, it must be that

$\sqrt{\frac{36}{4}}=\frac{\sqrt{36}}{\sqrt{4}}$

The quotient property $\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$

This suggests that in general, if $x$ and $y$ are positive real numbers,

$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}},$       $y\ne 0$

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

CAUTION
It is extremely important to remember that

$\begin{array}{lllll}\sqrt{x+y}\ne \sqrt{x}+\sqrt{y}\hfill & \hfill & \text{or}\hfill & \hfill & \sqrt{x-y}\ne \sqrt{x}-\sqrt{y}\hfill \end{array}$

For example, notice that $\sqrt{16+9}=\sqrt{25}=5,$ but $\sqrt{16}+\sqrt{9}=4+3=7.$

We shall study the process of simplifying a square root expression by distinguishing between two types of square roots: square roots not involving a fraction and square roots involving a fraction.

Square roots not involving fractions

A square root that does not involve fractions is in simplified form if there are no perfect square in the radicand.

The square roots $\sqrt{x,}\sqrt{ab},\sqrt{5mn,}\sqrt{2\left(a+5\right)}$ are in simplified form since none of the radicands contains a perfect square.

The square roots $\sqrt{{x}^{2},}\sqrt{{a}^{3}}=\sqrt{{a}^{2}a}$ are not in simplified form since each radicand contains a perfect square.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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