7.1 Division of square root expressions

Algebra ii for the community Page 1 / 1

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.

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

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

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.

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

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

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

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

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

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

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

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

$\frac{\sqrt{5 a}}{\sqrt{b}} = \frac{\sqrt{5 a}}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{\sqrt{5 a} \sqrt{b}}{b} = \frac{\sqrt{5 a b}}{b}$

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

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

Practice set a

Simplify the square root expressions.

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

$\sqrt{2}$

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

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

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

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

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

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

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

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

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

$\sqrt{a - 4}$

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

$x^{n}$

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

$a^{m - 2}$

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

a - b

. Similarly, a conjugate of

a - b

a + b

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

$(a + b) (a - b) = a^{2} - a b + a b - b^{2} = a^{2} - b^{2}$

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

$\begin{array}{l} (5 + \sqrt{2}) (5 - \sqrt{2}) & = & 5^{2} - {(\sqrt{2})}^{2} \\ = & 25 - 2 \\ = & 23 \end{array}$

$\begin{array}{l} (\sqrt{6} - \sqrt{7}) (\sqrt{6} + \sqrt{7}) & = & {(\sqrt{6})}^{2} - {(\sqrt{7})}^{2} \\ = & 6 - 7 \\ = & - 1 \end{array}$

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 $\frac{4 - \sqrt{6}}{4 - \sqrt{6}}$ . $\begin{array}{l} \frac{3}{4 + \sqrt{6}} \cdot \frac{4 - \sqrt{6}}{4 - \sqrt{6}} & = & \frac{3 (4 - \sqrt{6})}{4^{2} - {(\sqrt{6})}^{2}} \\ = & \frac{12 - 3 \sqrt{6}}{16 - 6} \\ = & \frac{12 - 3 \sqrt{6}}{10} \end{array}$

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

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

$\begin{array}{l} \frac{\sqrt{2 x}}{\sqrt{3} - \sqrt{5 x}} \cdot \frac{\sqrt{3} + \sqrt{5 x}}{\sqrt{3} + \sqrt{5 x}} & = & \frac{\sqrt{2 x} (\sqrt{3} + \sqrt{5 x})}{{(\sqrt{3})}^{2} - {(\sqrt{5 x})}^{2}} \\ = & \frac{\sqrt{2 x} \sqrt{3} + \sqrt{2 x} \sqrt{5 x}}{3 - 5 x} \\ = & \frac{\sqrt{6 x} + \sqrt{10 x^{2}}}{3 - 5 x} \\ = & \frac{\sqrt{6 x} + x \sqrt{10}}{3 - 5 x} \end{array}$

Practice set b

Simplify the following expressions.

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

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

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

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

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

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

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

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

Exercises

For the following problems, simplify each expressions.

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

$\sqrt{14}$

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

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

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

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

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

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

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

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

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

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

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

$\frac{7}{15}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$6 m^{2}$

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

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

$5 y^{3}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\sqrt{x - 6}$

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

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

$\sqrt{x - 1}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$3 + \sqrt{6}$

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

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

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

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

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

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

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

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

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

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

Exercises for review

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

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

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

( [link] ) Construct the graph of $y = \frac{2}{3} x - 4.$
An xy-plane with gridlines, labeled negative five and five on the both axes.

A graph of a line passing through two points with coordinates three, negative two; and zero, negative five.

( [link] ) The symbol $\sqrt{x}$ represents which square root of the number $x, x \geq 0$ ?

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

$a + 4$

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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