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Roots and Radicals: (lesson 3 of 3)

## Multiplying and Dividing Radical Expressions

We multiply binomial expressions involving radicals by using the FOIL (First, Outer, Inner, Last) method.

Example 1: Multiply each of the following

\begin{aligned} \text{ a) } & \left( \sqrt{5} - 3 \right) \cdot \left( \sqrt{2} + 2 \right) \\ \text{ b) } & \left( 2 - 3 \sqrt{5} \right) \cdot \left( \sqrt{15} + 2 \sqrt{3} \right) \end{aligned}

Solution

\begin{aligned} \text{a)} \left( \sqrt{5} - 3 \right) \cdot \left( \sqrt{2} + 2 \right) & = \underbrace{\sqrt{5} \cdot \sqrt{2}}_\text{FIRST} + \underbrace{\sqrt{5}\cdot 2}_\text{OUTER} - \underbrace{3 \cdot \sqrt{2}}_\text{INNER} - \underbrace{3 \cdot 2}_\text{LAST} = \\\\ & = \sqrt{10} + 2\,\sqrt{5} - 3\,\sqrt{2} - 6 \end{aligned}
\begin{aligned} \text{a)} \left( 2 - 3\sqrt{5} \right) \cdot \left( \sqrt{15} + 2\sqrt{3} \right) & = \underbrace{2\cdot\sqrt{15}}_\text{FIRST} + \underbrace{2 \cdot 2 \sqrt{3}}_\text{OUTER} - \underbrace{3\sqrt{5} \cdot \sqrt{15}}_\text{INNER} - \underbrace{3\sqrt{5} \cdot 2\sqrt{3}}_\text{LAST} = \\\\ & = \color{blue}{2\sqrt{15}} + 4\sqrt{3} - 3\sqrt{\color{red}{75}} \color{blue}{- 6\sqrt{15}} = \\\\ & = \color{blue}{-4\sqrt{15}} + 4\sqrt{3} - 3\sqrt{\color{red}{25\cdot 3}} = \\\\ & = -4\sqrt{15} + 4\sqrt{3} - 3 \cdot \color{red}{5 \sqrt{3}} = \\\\ & = -4\sqrt{15} + 4\sqrt{3} -15 \sqrt{3} = \\\\ & = -4\sqrt{15} -11\sqrt{3} \end{aligned}

Example 2: Multiply each of the following

\begin{aligned} \text{ a) } & \left( \sqrt{7} - \sqrt{5} \right) \cdot \left( \sqrt{7} + \sqrt{5} \right) \\\\ \text{ b) } & \left( \sqrt{12} - 2 \right)^2 \end{aligned}

Solution

\begin{aligned} \text{a)} \left( \sqrt{7} - \sqrt{5} \right) \cdot \left( \sqrt{7} + \sqrt{5} \right) & = \underbrace{\sqrt{7} \cdot \sqrt{7}}_\text{FIRST} + \underbrace{\sqrt{7}\cdot \sqrt{5}}_\text{OUTER} - \underbrace{\sqrt{5} \cdot \sqrt{7}}_\text{INNER} - \underbrace{\sqrt{5} \cdot \sqrt{5}}_\text{LAST} = \\\\ & = \sqrt{49} + \cancel{\color{red}{\sqrt{35}}} - \cancel{\color{red}{\sqrt{35}}} - \sqrt{25} = \\\\ &= 7 - 5 = 2 \end{aligned}
\begin{aligned} \text{b)} \left( \sqrt{12} - 2 \right)^2 & = \left( \sqrt{12} - 2 \right) \cdot \left( \sqrt{12} - 2 \right) = \\\\ &= \underbrace{\sqrt{12} \cdot \sqrt{12}}_\text{FIRST} - \underbrace{\sqrt{12}\cdot 2}_\text{OUTER} - \underbrace{2 \cdot \sqrt{12}}_\text{INNER} + \underbrace{2 \cdot 2}_\text{LAST} = \\\\ & = \sqrt{144} - 2\sqrt{12} - 2\sqrt{12} + 4 = \\\\ &= 12 - 4\sqrt{12} + 4 = \\\\ &= 16 - 4\sqrt{12} \end{aligned}

Exercise 1: Multiply each of the following

Level 1

 $$\color{blue}{\left( \sqrt5 - \sqrt3 \right) \left( \sqrt3 - \sqrt5\right) = }$$ $-2$ $2$ $4$ $-8$

Level 2

 $$\color{blue}{\left( 2 \sqrt5 - \sqrt2 \right)^2 = }$$ $2\sqrt{10} - 2$ $\sqrt{10} - 11$ $22 - 2\sqrt{10}$ $18$

A common way of dividing the radical expression is to have the denominator that contain no radicals. Dividing radical is based on rationalizing the denominator. Rationalizing is the process of starting with a fraction containing a radical in its denominator and determining fraction with no radical in its denominator.

Techniques for rationalizing the denominator are shown below.

CASE 1: Rationalizing denominators with one square roots

When you have one root in the denominator you multiply top and bottom by it.

Example 3: Rationalize each denominator

$$\text{a) } \frac{10}{\sqrt{5}} \qquad \text{b) } \frac{4}{\sqrt{8}}$$

Solution

$$\text{a) } \frac{10}{\sqrt{5}} = \frac{10}{\sqrt{5}} \cdot \frac{\color{blue}{\sqrt{5}}}{\color{blue}{\sqrt{5}}} = \frac{ 10 \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{ 10 \sqrt{5}}{\sqrt{25}} = \frac{ 10 \sqrt{5}}{5} = 2\sqrt{5}$$
$$\text{b) } \frac{4}{\sqrt{8}} = \frac{4}{\sqrt{8}} \cdot \frac{\color{blue}{\sqrt{8}}}{\color{blue}{\sqrt{8}}} = \frac{ 4 \sqrt{8}}{\sqrt{8} \cdot \sqrt{8}} = \frac{ 4 \sqrt{4\cdot 2}}{\sqrt{64}} = \frac{ 4 \cdot 2 \cdot \sqrt{2}}{8} = \frac{8\sqrt{2}}{8} = \sqrt{2}$$

Exercise 2: Rationalize each denominator

Level 1

 $$\color{blue}{\frac{12}{\sqrt3}}$$ $12\sqrt3$ $6\sqrt3$ $4\sqrt3$ $\sqrt3$

Level 2

 $$\color{blue}{\frac{24}{\sqrt{12}}}$$ $\sqrt{12}$ $4\sqrt{3}$ $2\sqrt{24}$ $\sqrt{2}$

CASE 2: Rationalizing Denominators with Cube Roots

Here you need to multiply the numerator and denominator by a number that will result in a perfect cube in the radicand in the denominator.

Example 4: Rationalize each denominator

$$\text{a) } \frac{10}{\sqrt[\large{3}]{2}} \qquad \text{b) } \frac{18}{\sqrt[\large{3}]{9}}$$

Solution

$$\text{b) } \frac{10}{\sqrt[\large{3}]{2}} = \frac{10}{\sqrt[\large{3}]{2}} \cdot \frac{\color{blue}{\sqrt[\large{3}]{4}}}{\color{blue}{\sqrt[\large{3}]{4}}} = \frac{ 10 \sqrt[\large{3}]{4}}{\sqrt[\large{3}]{2} \cdot \sqrt[\large{3}]{4}} = \frac{ 10 \sqrt[\large{3}]{4}}{\sqrt[\large{3}]{8}} = \frac{ 10 \sqrt[\large{3}]{4}}{2} = 5\sqrt[\large{3}]{2}$$
$$\text{a) } \frac{18}{\sqrt[\large{3}]{9}} = \frac{18}{\sqrt[\large{3}]{9}} \cdot \frac{\color{blue}{\sqrt[\large{3}]{3}}}{\color{blue}{\sqrt[\large{3}]{3}}} = \frac{18 \sqrt[\large{3}]{9}}{\sqrt[\large{3}]{9} \cdot \sqrt[\large{3}]{3}} = \frac{18 \sqrt[\large{3}]{9}}{\sqrt[\large{3}]{27}} = \frac{18 \sqrt[\large{3}]{9}}{3} = 6\sqrt[\large{3}]{9}$$

Exercise 3: Rationalize each denominator

Level 1

 $$\color{blue}{\frac{8}{\sqrt[\large{3}]{2}}}$$ $4 \sqrt[\large{3}]{4}$ $4 \sqrt[\large{3}]{2}$ $2 \sqrt[\large{3}]{4}$ $2 \sqrt[\large{3}]{2}$

Level 2

 $$\color{blue}{\frac{8}{\sqrt[\large{3}]{4}}}$$ $4 \sqrt[\large{3}]{2}$ $4 \sqrt[\large{3}]{4}$ $2 \sqrt[\large{3}]{4}$ $2 \sqrt[\large{3}]{2}$

CASE 3: Rationalize denominators with binomials

In this case, you will need to multiply the denominator and numerator by the same expression as the denominator but with the opposite sign in the middle. This expression is called the conjugate of the denominator.

Example 5: Rationalize denominator in $\frac{4}{\sqrt{5} + \sqrt{3}}$ .

In this example denominator is $\color{blue}{\sqrt{5}} \color{red}{+} \color{blue}{\sqrt{3}}$ so we will multiply the denominator and numerator with $\color{blue}{\sqrt{5}} \color{red}{-} \color{blue}{\sqrt{3}}$. Here is a complete solution:

\begin{aligned} \frac{4}{\sqrt{5} + \sqrt{3}} &= \frac{4}{\sqrt{5} \color{red}{+} \sqrt{3}} \cdot \frac{ \sqrt{5} \color{red}{-} \sqrt{3} }{ \sqrt{5} \color{red}{-} \sqrt{3} } =\\\\ &= \frac{ 4 \left( \sqrt{5} - \sqrt{3} \right) }{\left( \sqrt{5} + \sqrt{3} \right)\left( \sqrt{5} - \sqrt{3} \right)} = \\\\ &= \frac{ 4 \left( \sqrt{5} - \sqrt{3} \right) }{\sqrt{5}\cdot\sqrt{5} - \sqrt{5} \cdot \sqrt{3} + \sqrt{3} \cdot \sqrt{5} - \sqrt{3}\cdot \sqrt{3}} = \\\\ &= \frac{ 4 \left( \sqrt{5} - \sqrt{3} \right) }{\sqrt{25} - \cancel{\color{blue}{\sqrt{15}}} + \cancel{\color{blue}{\sqrt{15}}} - \sqrt{9}} = \\\\ &= \frac{ 4 \left( \sqrt{5} - \sqrt{3} \right) }{5 - 3} = \\\\ &= \frac{ \cancel{\color{blue}{4}} \left( \sqrt{5} - \sqrt{3} \right) }{\cancel{\color{blue}{2}}} = \\\\ &= 2 \left( \sqrt{5} - \sqrt{3} \right) \end{aligned}

Example 6: Rationalize denominator in $\frac{\sqrt 8 - \sqrt 3}{\sqrt 6 - 2}$ .

In this example denominator is $\sqrt 6 \color{red}{-} 2$ so we will multiply the denominator and numerator by $\sqrt 6 \color{red}{+} 2$. The solution is:

\begin{aligned} \frac{\sqrt 8 - \sqrt 3}{\sqrt 6 - 2} &= \frac{\sqrt 8 - \sqrt 3}{\sqrt 6 - 2} \cdot \frac{\color{blue}{\sqrt 6 + 2}}{\color{blue}{\sqrt 6 + 2}} = \\\\ &= \frac{\left(\sqrt 8 - \sqrt3\right)\cdot\left(\sqrt 6 + 2\right)}{\left(\sqrt 6 - \sqrt 2\right) \cdot \left(\sqrt 6 + \sqrt 2\right) } = \\\\ &= \frac{\sqrt 8 \cdot \sqrt 6 + 2 \sqrt 8 - \sqrt 3 \cdot \sqrt 6 - 2 \sqrt 3 }{\sqrt 6 \cdot \sqrt 6 + 2 \sqrt 6 - 2 \sqrt 6 - 2 \cdot 2 } = \\\\ &= \frac{\sqrt{48} + 2\sqrt 8 - \sqrt{18} - 2 \sqrt 3}{\sqrt{36} - 2\sqrt 6 + 2\sqrt 6 - 4} = \\\\ &= \frac{4 \sqrt 3 + 2\cdot 2 \sqrt2 - 3\sqrt2 - 2\sqrt3}{6-4} = \\\\ &= \frac{2 \sqrt3 - \sqrt2}{2} \end{aligned}

Exercise 4: Rationalize each denominator

Level 1

 $$\color{blue}{\frac{2}{\sqrt5 - \sqrt3}}$$ $\sqrt5 - \sqrt3$ $\sqrt5 + \sqrt3$ $- \sqrt5 + \sqrt3$ $- \sqrt5 - \sqrt3$

Level 2

 $$\color{blue}{\frac{\sqrt{18}-\sqrt2 }{\sqrt7 - 2}}$$ $2\left(\sqrt7 - \sqrt5 \right)$ $2\left(\sqrt7 + \sqrt5 \right)$ $2\sqrt2\left(\sqrt7 - \sqrt5 \right)$ $\sqrt2\left(\sqrt7 - \sqrt5 \right)$