Q. 71 - Intermediate Algebra

Question

In the following exercises, simplify using absolute value signs as needed.$\sqrt{125r^{13}},\sqrt[3]{108x^5},\sqrt[4]{48y^6}$

Step-by-Step Solution

Answer

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

Step 1: Given information

Given the expression

Step 2: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt{125r^{13}}=\sqrt{25r^{12}·5r^1} \\ \sqrt{125r^{13}}=\sqrt{{\left(5r^6\right)}^2·5r} \\ \sqrt{125r^{13}}=\sqrt{{\left(5r^6\right)}^2}·\sqrt{5r} \\ \sqrt{125r^{13}}=5\left|r^6\right|\sqrt{5r} \\ \sqrt{125r^{13}}=5r^6\sqrt{5r} \end{gathered}$$

Step 3: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt[3]{108x^5}=\sqrt[3]{27x^3·4x^2} \\ \sqrt[3]{108x^5}=\sqrt[3]{(3x{)}^3}·\sqrt[3]{4x^2} \\ \sqrt[3]{108x^5}=3x\sqrt[3]{4x^2} \end{gathered}$$

Step 4: Using the property

Using the property, we get

$$\begin{gathered} \sqrt[4]{48y^6}=\sqrt[4]{16y^4·3y^2} \\ \sqrt[4]{48y^6}=\sqrt[4]{16y^4·3y^2} \\ \sqrt[4]{48y^6}=\sqrt[4]{(2y{)}^4}·\sqrt[4]{3y^2} \\ \sqrt[4]{48y^6}=2\left|y\right|\sqrt[4]{3y^2} \end{gathered}$$