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Roots and Radicals: (lesson 2 of 3)

Step 1. Simplify radicals. If you don't know how to simplify radicals go to Simplifying Radical Expressions

Example 1: Add or subtract to simplify radical expression: $2 \sqrt{12} + \sqrt{27}$

Solution:

\begin{aligned} \sqrt{12} &= \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2 \sqrt{3}\\ \sqrt{27} &= \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3 \sqrt{3} \end{aligned}

$$2 \color{red}{\sqrt{12}} + \color{blue}{\sqrt{27}} = 2\cdot \color{red}{2 \sqrt{3}} + \color{blue}{3\sqrt{3}} = \underbrace{ 4\sqrt{3} + 3\sqrt{3} = 7\sqrt{3}}_\text{COMBINE LIKE TERMS}$$

Example 2: Add or subtract to simplify radical expression: $3 \sqrt{50} - 2 \sqrt{8} - 5 \sqrt{32}$

Solution:

\begin{aligned} \sqrt{50} &= \sqrt{25 \cdot 2} = 5 \sqrt{2} \\ \sqrt{8} &= \sqrt{4 \cdot 2} = 2 \sqrt{2} \\ \sqrt{32} &= \sqrt{16 \cdot 2} = 4 \sqrt{2} \end{aligned}

\begin{aligned} 3 \color{red}{\sqrt{50}} - 2 \color{blue}{\sqrt{8}} - 5 \color{green}{\sqrt{32}} &= \\ &= 3 \cdot \color{red}{5 \sqrt{2}} - 2 \cdot \color{blue}{2 \sqrt{2}} - 5 \cdot \color{green}{4 \sqrt{2}} = \\ &= \underbrace{ 15 \sqrt{2} - 4 \sqrt{2} - 20 \sqrt{2} = -9 \sqrt{2}}_\text{COMBINE LIKE TERMS} \end{aligned}

Exercise 1: Add or subtract to simplify

Level 1

 $$\color{blue}{\sqrt{50} - \sqrt{32} = }$$ $-\sqrt{3}$ $-\sqrt{2} + \sqrt{3}$ $-\sqrt{2}$ $\sqrt{2}$

Level 2

 $$\color{blue}{2\sqrt{12} - 3 \sqrt{27}}$$ $-5\sqrt{3}$ $3\sqrt{3}$ $4\sqrt{2}$ $-4\sqrt{2}$

Example 3: Add or subtract to simplify radical expression: $4 \sqrt{2} - 3 \sqrt{3}$

Solution:

Here the radicands differ and are already simplified, so this expression cannot be simplified.

Example 4: Add or subtract to simplify radical expression: $4 \sqrt{ \frac{20}{9} } + 5 \sqrt{ \frac{45}{16} }$

Solution:

\begin{aligned} \color{blue}{\sqrt{ \frac{20}{9} }} &= \frac{\sqrt{20}}{\sqrt{9}} = \frac{\sqrt{4 \cdot 5}}{3} = \frac{2 \cdot \sqrt{5}}{3} = \color{blue}{\frac{2}{3} \cdot \sqrt{5}} \\ \color{red}{\sqrt{ \frac{45}{16} }} &= \frac{\sqrt{45}}{\sqrt{16}} = \frac{\sqrt{9 \cdot 5}}{4} = \frac{3 \cdot \sqrt{5}}{4} = \color{red}{\frac{3}{4} \cdot \sqrt{5}} \\ \end{aligned}

\begin{aligned} 4 \cdot \color{blue}{\sqrt{\frac{20}{9}}} + 5 \cdot \color{red}{\sqrt{\frac{45}{16}}} &= \\ &= 4 \cdot \color{blue}{\frac{2}{3} \cdot \sqrt{5}} + 5 \cdot \color{red}{\frac{3}{4} \cdot \sqrt{5}} = \\ &= \frac{8}{3} \cdot \sqrt{5} + \frac{15}{4} \cdot \sqrt{5} = \\ &= \left( \frac{8}{3} + \frac{15}{4} \right) \sqrt{5} = \frac{77}{12} \sqrt{5} \end{aligned}

Example 5: Add or subtract to simplify radical expression: $6 \sqrt{ \frac{24}{x^4}} - 3 \sqrt{ \frac{54}{x^4}}$

Solution:

\begin{aligned} \color{blue}{\sqrt{\frac{24}{x^4}}} &= \frac{\sqrt{24}}{\sqrt{x^4}} = \frac{\sqrt{4 \cdot 6}}{x^2} = \color{blue}{\frac{2 \sqrt{6}}{x^2}} \\ \color{red}{\sqrt{\frac{54}{x^4}}} &= \frac{\sqrt{54}}{\sqrt{x^4}} = \frac{\sqrt{9 \cdot 6}}{x^2} = \color{red}{\frac{3 \sqrt{6}}{x^2}} \end{aligned}

$$6 \cdot \color{blue}{\sqrt{\frac{24}{x^4}}} - 3 \cdot \color{red}{\sqrt{\frac{54}{x^4}}} = 6 \cdot \color{blue}{\frac{2 \sqrt{6}}{x^2}} - 3 \cdot \color{red}{\frac{3 \sqrt{6}}{x^2}} = \frac{12 \sqrt{6}}{x^2} - \frac{9 \sqrt{6}}{x^2} = \frac{12 \sqrt{6} - 9 \sqrt{6}}{x^2} = \frac{3\sqrt{6}}{x^2}$$
 $$\color{blue}{4\sqrt{\frac{3}{4}} + 8 \sqrt{ \frac{27}{16}} }$$ $-2\sqrt{2}$ $-3\sqrt{2}$ $8\sqrt{2}$ $-4\sqrt{2}$
 $$\color{blue}{ 3\sqrt{\frac{3}{a^2}} - 2 \sqrt{\frac{12}{a^2}}}$$ $\frac{\sqrt3}{a}$ $-\frac{\sqrt3}{a}$ $-a\sqrt3$ $a\sqrt3$