Q. 72 - Intermediate Algebra

Question

In the following exercises, simplify using absolute value signs as needed. $\sqrt{80s^{15}},\sqrt[5]{96a^7},\sqrt[6]{128b^7}$

Step-by-Step Solution

Answer

The expression in simplified form is $4s^6\sqrt{5s^3},2a\sqrt[5]{3a^2},2\left|b\right|\sqrt[6]{2b}$ respectively

Step 1: Given information

Given the expression $\sqrt{80s^{15}},\sqrt[5]{96a^7},\sqrt[6]{128b^7}$

Step 2: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt{80s^{15}}=\sqrt{16s^{12}·5s^3} \\ \sqrt{80s^{15}}=\sqrt{{\left(4s^6\right)}^2.5s^3} \\ \sqrt{80s^{15}}=\sqrt{{\left(4s^6\right)}^2}·\sqrt{5s^3} \\ \sqrt{80s^{15}}==4\left|s^6\right|\sqrt{5s^3} \\ \sqrt{80s^{15}}=4s^6\sqrt{5s^3} \end{gathered}$$

Step 3: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt[5]{96a^7}=\sqrt[5]{32a^5·3a^2} \\ \sqrt[5]{96a^7}=\sqrt[5]{(2a{)}^5}·\sqrt[5]{3a^2} \\ \sqrt[5]{96a^7}=2a\sqrt[5]{3a^2} \end{gathered}$$

Step 4: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt[6]{128b^7}=\sqrt[6]{64b^6·2b} \\ \sqrt[6]{128b^7}=\sqrt[6]{(2b{)}^6·2b} \\ \sqrt[6]{128b^7}=\sqrt[6]{(2b{)}^6}·\sqrt[6]{2b} \\ \sqrt[6]{128b^7}=2\left|b\right|\sqrt[6]{2b} \end{gathered}$$