Q. 8.48 - Intermediate Algebra

Question

Simplify:

$\left(a\right) \sqrt{\frac{48m^7{n}^2}{100m^5{n}^8} } \left(b\right) \sqrt[3]{\frac{54x^7{y}^5}{250x^2{y}^2}} \left(c\right) \sqrt[4]{\frac{32a^9{b}^7}{162a^3{b}^3}}.$

Step-by-Step Solution

Answer

Part a. After simplifying we get, $\frac{2m\sqrt{3}}{5n^3}.$

Part b. After simplifying we get, $\frac{3xy \sqrt[3]{x^2}}{5}.$

Part c. After simplifying we get, $\frac{2ab \sqrt[4]{a^2}}{3}.$

Part a Step 1. Simplify the fraction in the radicand

Given: $\sqrt{\frac{48m^7{n}^2}{100m^5{n}^8}}$

After simplifying the fraction in the radicand we get,

$$\begin{gathered} \sqrt{\frac{48m^{7-5}{n}^{2-8}}{100}} \\ =\sqrt{\frac{12m^2{n}^{-6}}{25}} \\ =\sqrt{\frac{12m^2}{25n^6}} \end{gathered}$$

Part a Step 2. Use the Quotient property to rewrite the radical as the quotient of two radicals

So, we get $\frac{\sqrt{12m^2}}{\sqrt{25n^6}}.$

Part a Step 3. Simplify the radicals of both the numerator and denominator

After simplifying we get, 

$$\begin{gathered} \frac{\sqrt{12m^2}}{\sqrt{25n^6}} \\ =\frac{\sqrt{{\left(2m\right)}^2}·\sqrt{3}}{\sqrt{{\left(5n^3\right)}^2}} \\ =\frac{2m\sqrt{3}}{5n^3} \end{gathered}$$

Part b Step 1. Simplify the fraction in the radicand

Given: $\sqrt[3]{\frac{54x^7{y}^5}{250x^2{y}^2}}$

After simplifying the fraction in the radicand we get,

$$\begin{gathered} \sqrt[3]{\frac{27x^{7-2}{y}^{5-2}}{125}} \\ =\sqrt[3]{\frac{27x^5{y}^3}{125}} \end{gathered}$$

Part b Step 2. Use the Quotient property to rewrite the radical as the quotient of two radicals

So, we get $\frac{\sqrt[3]{27x^5{y}^3}}{\sqrt[3]{125}}.$

Part b Step 3. Simplify the radicals of both the numerator and denominator

After simplifying we get,  

$$\begin{gathered} \frac{\sqrt[3]{27x^5{y}^3}}{\sqrt[3]{125}} \\ =\frac{\sqrt[3]{{\left(3xy\right)}^3}·\sqrt[3]{x^2}}{\sqrt[3]{{\left(5\right)}^3}} \\ =\frac{3xy \sqrt[3]{x^2}}{5} \end{gathered}$$

Part c Step 1. Simplify the fraction in the radicand

Given:  $\sqrt[4]{\frac{32a^9{b}^7}{162a^3{b}^3}}$

After simplifying the fraction in the radicand we get,

$$\begin{gathered} \sqrt[4]{\frac{16a^{9-3}{b}^{7-3}}{81}} \\ =\sqrt[4]{\frac{16a^6{b}^5}{81}} \end{gathered}$$

Part c Step 2. Use the Quotient property to rewrite the radical as the quotient of two radicals

So, we get $\frac{\sqrt[4]{16a^6{b}^4}}{\sqrt[4]{81}}.$

Part c Step 3. Simplify the radicals of both the numerator and denominator

After simplifying we get, 

$$\begin{gathered} \frac{\sqrt[4]{16a^6{b}^4}}{\sqrt[4]{81}} \\ =\frac{\sqrt[4]{{\left(2ab\right)}^4}·\sqrt[4]{a^2}}{\sqrt[4]{{\left(3\right)}^4}} \\ =\frac{2ab \sqrt[4]{a^2}}{3} \end{gathered}$$