Q. 8.47 - Intermediate Algebra

Question

Simplify:

$\left(a\right) \sqrt{\frac{50x^5{y}^3}{72x^{4}y}} \left(b\right) \sqrt[3]{\frac{16x^5{y}^7}{54x^2{y}^2}} \left(c\right) \sqrt[4]{\frac{5a^8{b}^6}{80a^3{b}^2}}.$

Step-by-Step Solution

Answer

Part a. After simplifying we get, $\frac{5y\sqrt{x}}{6}.$

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

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

Part a Step 1. Simplify the fraction in the radicand

Given: $\sqrt{\frac{50x^5{y}^3}{72x^{4}y}}$

After simplifying the fraction in the radicand we get,

 $$\begin{gathered} \sqrt{\frac{25{\left(x\right)}^{5-4}{\left(y\right)}^{3-1}}{36}} \\ =\sqrt{\frac{25x{y}^2}{36}} \end{gathered}$$

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

So, we get $\frac{\sqrt{25x{y}^2}}{\sqrt{36}}.$

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

After simplifying we get,

$$\begin{gathered} \frac{\sqrt{25x{y}^2}}{\sqrt{36}} \\ =\frac{\sqrt{{\left(5y\right)}^{2}x}}{\sqrt{{\left(6\right)}^2}} \\ =\frac{5y\sqrt{x}}{6} \end{gathered}$$

Part b Step 1. Simplify the fraction in the radicand

Given: $\sqrt[3]{\frac{16x^5{y}^7}{54x^2{y}^2}}$

After simplifying the fraction in the radicand we get,

 $$\begin{gathered} \sqrt[3]{\frac{8{\left(x\right)}^{5-2}{\left(y\right)}^{7-2}}{27}} \\ =\sqrt[3]{\frac{8x^3{y}^5}{27}} \end{gathered}$$

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

So, we get $\frac{\sqrt[3]{8x^3{y}^5}}{\sqrt[3]{27}}.$

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

After simplifying we get, 

$$\begin{gathered} \frac{\sqrt[3]{8x^3{y}^5}}{\sqrt[3]{27}} \\ =\frac{\sqrt[3]{{\left(2xy\right)}^3}·\sqrt[3]{y^2}}{\sqrt[3]{{\left(3\right)}^3}} \\ =\frac{2xy \sqrt[3]{y^2}}{3} \end{gathered}$$

Part c Step 1. Simplify the fraction in the radicand

Given:  $\sqrt[4]{\frac{5a^8{b}^6}{80a^3{b}^2}}$

After simplifying the fraction in the radicand we get,

$$\begin{gathered} \sqrt[4]{\frac{a^{8-3}{b}^{6-2}}{16}} \\ =\sqrt[4]{\frac{a^5{b}^4}{16}} \end{gathered}$$

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

So, we get $\frac{\sqrt[4]{a^5{b}^4}}{\sqrt[4]{16}}.$

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

After simplifying we get, 

$$\begin{gathered} \frac{\sqrt[4]{a^5{b}^4}}{\sqrt[4]{16}} \\ =\frac{\left|ab\right|\sqrt[4]{a}}{2} \end{gathered}$$