407 493 6601

# Algebra 1 - Multiplying and Dividing Radical Expressions

## Introduction

• In multiplying radical expressions, we use the Product Rule: If  represents nonnegative real numbers, then $\sqrt{x}·\sqrt{y}=\sqrt{xy}$.
• For higher indices, we have  are nonnegative real numbers.
• In dividing radical expressions, the Quotient Rule holds: Given that . If n is an even integer, then both x and y are nonnegatives.

Important Key Points:

• We can multiply radical expressions having the same index yet different radicands. In some cases, after multiplying, we need to simplify the result.
• We can also multiply two or more radicals with different indices by following a series of steps.
• Certain algebraic rules in multiplying polynomials (i.e. special product cases) can apply to the multiplication of radical expressions.
• In the division of radicals with two terms in the denominator, we can rationalize this type of radical expression by multiplying the numerator and denominator by the conjugate of the denominator.

## Solved Examples

Example 1. Find the product: $7xy\sqrt{2x}·4{z}^{3}\sqrt{8{x}^{3}}$

Solution:

Example 2. Find the product: $6\sqrt{5a{b}^{3}}·7\sqrt{25{a}^{4}{b}^{6}}$

Solution:

Example 3. What is the product of ?

Solution:

$\sqrt{7}·\sqrt{5}={\left(7\right)}^{\frac{1}{2}}{\left(5\right)}^{\frac{1}{3}}={\left(7\right)}^{\frac{3}{6}}{\left(5\right)}^{\frac{2}{6}}={\left({7}^{3}·{5}^{2}\right)}^{\frac{1}{6}}={\left(8575\right)}^{\frac{1}{6}}=\sqrt{8575}$

Example 4. Find the product: $\left(\sqrt{7x}-\sqrt{3y}\right)\left(\sqrt{7x}+\sqrt{3y}\right)$

Solution:

$\left(\sqrt{7x}-\sqrt{3y}\right)\left(\sqrt{7x}+\sqrt{3y}\right)=\sqrt{49{x}^{2}}+\sqrt{21xy}-\sqrt{21xy}-\sqrt{9{y}^{2}}=7x-3y$

Example 5. Give the product: ${\left(\sqrt{5a}-b\sqrt{7c}\right)}^{2}$

Solution:

${\left(\sqrt{5a}-b\sqrt{7c}\right)}^{2}={\left(\sqrt{5a}\right)}^{2}-2\left(\sqrt{5a}\right)\left(b\sqrt{7c}\right)+{\left(b\sqrt{7c}\right)}^{2}=5a-2b\sqrt{35ac}+7{b}^{2}c$

Example 6. Find the quotient: $\frac{17\sqrt{28}+3\sqrt{14}-2\sqrt{98}}{\sqrt{7}}$

Solution:

$\frac{17\sqrt{28}}{\sqrt{7}}+\frac{3\sqrt{14}}{\sqrt{7}}-\frac{2\sqrt{98}}{\sqrt{7}}=17\sqrt{4}+3\sqrt{2}-2\sqrt{14}=34+3\sqrt{2}-2\sqrt{14}$

Example 7. Simplify the expression $\frac{7\sqrt{3}+\sqrt{7}}{7\sqrt{3}-\sqrt{7}}$

Solution:

## Cheat Sheet

• To simplify the expression $\sqrt{\frac{x-y}{x+y}}$, Quotient Rule is applicable where .
• The conjugate of the denominator of $\frac{\sqrt{11}-\sqrt{3x}}{2\sqrt{7x}+3\sqrt{6}}$ is $2\sqrt{7x}-3\sqrt{6}$.
• To simplify the expression $\frac{4}{\sqrt{x}+y}$, rationalize the denominator by multiplying the denominator by its conjugate which is . This is based on the concept of $\left(x+y\right)\left({x}^{2}-xy+{y}^{2}\right)={x}^{3}+{y}^{3}$.
• When dividing two radicals of the same order, use the Quotient Rule and siimplfy it by rationalizing the denominator.
• In multiplying two finite nested radicals like $\sqrt{7-3\sqrt{6}}·\sqrt{7+3\sqrt{6}}$, we treate  as radicands.
• To rationalize $\frac{7}{\sqrt{x}}$, multiply both numerator and denominator by $\sqrt{{x}^{4}}$.

## Blunder Areas

• The process of multiplying the square root of two negative numbers such as $\sqrt{-16}·\sqrt{-9}=\sqrt{144}$ is incorrect since the radicand must be nonnegative real numbers. $\sqrt{-16}=4i$ and $\sqrt{-9}=3i$.
• The square root of negative numbers is not defined under the real number system.
• When multiplying and dividing radical expressions, it is imperative to simplify the final result whenever possible.