Chapter 7.4, Problem 44PPS

To determine

To simplify the expression $\sqrt{\frac{3}{16}}\cdot \sqrt{\frac{9}{5}}$ .

Answer to Problem 44PPS

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

Explanation of Solution

Given:

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

Formula used:

The radical rules:

  $\sqrt[n]{a}=a^{\frac{1}{n}}$

  ${(a^m)}^n=a^{mn}$

  $\sqrt[a]{b\times c}=\sqrt[a]{b}\times \sqrt[a]{c}$

Calculation:

  $\begin{array}{l}\Rightarrow \sqrt{\frac{3}{16}}\cdot \sqrt{\frac{9}{5}}\\ \Rightarrow \sqrt{\frac{3}{2\times 2\times 2\times 2}}\times \sqrt{\frac{3\times 3}{5}}\\ \Rightarrow \sqrt{3\times 3\times 3}\times \frac{1}{\sqrt{2\times 2\times 2\times 2}}\times \frac{1}{\sqrt{5}}[\because \sqrt[a]{b\times c}=\sqrt[a]{b}\times \sqrt[a]{c}]\\ \Rightarrow \sqrt{3^3}\times \frac{1}{\sqrt{2^4}}\times \frac{1}{\sqrt{5}}\end{array}$

  $\begin{array}{l}$ \Rightarrow {(3^3)}^{\frac{1}{2}}\times \frac{1}{{\left(2^4\right)}^{\frac{1}{2}}}\times \frac{1}{{\left(5\right)}^{\frac{1}{2}}}[\because \sqrt[a]{b}={(b)}^{\frac{1}{a}}]$$ \Rightarrow {(3^3)}^{\frac{1}{2}}\times \frac{1}{{\left(2^4\right)}^{\frac{1}{2}}}\times \frac{1}{{\left(5\right)}^{\frac{1}{2}}}[\because \sqrt[a]{b}={(b)}^{\frac{1}{a}}]$\Rightarrow 3\times \sqrt{3}\times \frac{1}{2^2}\times \frac{1}{\sqrt{5}}\\ \Rightarrow \frac{3}{4}\cdot \frac{\sqrt{3}}{\sqrt{5}}\\ \Rightarrow \frac{3}{4}\cdot \sqrt{\frac{3}{5}}\end{array}$

Conclusion:

The simplified form of the given expression is equal to $\frac{3}{4}\cdot \sqrt{\frac{3}{5}}$ .