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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: $6\sqrt[3]{5a{b}^{3}}·7\sqrt[3]{25{a}^{4}{b}^{6}}$

Solution:

Example 3. What is the product of ?

Solution:

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

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 $\frac{4}{\sqrt[3]{x}+y}$, 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 $\frac{7}{\sqrt[5]{x}}$, multiply both numerator and denominator by $\sqrt[5]{{x}^{4}}$.

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.