# How do you simplify sqrt(3/16)*sqrt(9/5)?

Apr 23, 2017

$\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}} = \frac{3 \sqrt{15}}{20}$

#### Explanation:

Note that if $a , b \ge 0$ then:

$\sqrt{a b} = \sqrt{a} \sqrt{b}$

If $a > 0$ and $b \ge 0$ then:

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

If $a \ge 0$ then:

$\sqrt{{a}^{2}} = a$

When simplifying square roots of rational expressions, I like to make the denominator square before splitting the square root. That way we don't have to rationalise the denominator later...

$\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}} = \sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9 \cdot 5}{5 \cdot 5}}$

$\textcolor{w h i t e}{\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}}} = \sqrt{\frac{3}{4} ^ 2} \cdot \sqrt{\frac{45}{5} ^ 2}$

$\textcolor{w h i t e}{\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}}} = \sqrt{\frac{3 \cdot 45}{{4}^{2} \cdot {5}^{2}}}$

$\textcolor{w h i t e}{\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}}} = \frac{\sqrt{3 \cdot 45}}{\sqrt{{4}^{2} \cdot {5}^{2}}}$

$\textcolor{w h i t e}{\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}}} = \frac{\sqrt{3 \cdot {3}^{2} \cdot 5}}{\sqrt{{\left(4 \cdot 5\right)}^{2}}}$

$\textcolor{w h i t e}{\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}}} = \frac{\sqrt{{3}^{2}} \cdot \sqrt{15}}{\sqrt{{20}^{2}}}$

$\textcolor{w h i t e}{\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}}} = \frac{3 \sqrt{15}}{20}$

Apr 23, 2017

$\frac{3 \sqrt{15}}{20}$

#### Explanation:

$\sqrt{\frac{3}{16}} \cdot \sqrt{\frac{9}{5}}$

$\therefore = \frac{\sqrt{3}}{\sqrt{16}} \cdot \frac{\sqrt{9}}{\sqrt{5}}$

$\therefore = \frac{\sqrt{3}}{\sqrt{4 \cdot 4}} \cdot \frac{\sqrt{3 \cdot 3}}{\sqrt{5}}$

$\therefore \sqrt{4} \cdot \sqrt{4} = 4 , \sqrt{3} \cdot \sqrt{3} = 3$

$\therefore = \frac{\sqrt{3}}{4} \cdot \frac{3}{\sqrt{5}}$

$\therefore = \frac{3 \sqrt{3}}{4 \sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

$\therefore = \sqrt{5} \cdot \sqrt{5} = 5$

$\therefore = \frac{3 \sqrt{15}}{4 \cdot 5}$

$\therefore = \frac{3 \sqrt{15}}{20}$