Q. 75 - Intermediate Algebra

Question

In the following exercises, simplify using absolute value signs as needed.$\sqrt{147m^7{n}^{11}},\sqrt[3]{48x^6{y}^7},\sqrt[4]{32x^5{y}^4}$

Step-by-Step Solution

Answer

The expression on simplification is $7\left|m^3{n}^5\right|\sqrt{3mn},2x^2{y}^2\sqrt[3]{6y},2\left|xy\right|\sqrt[4]{2x}$ respectively

Step 1: Given information

Given the expression $\sqrt{147m^7{n}^{11}},\sqrt[3]{48x^6{y}^7},\sqrt[4]{32x^5{y}^4}$

Step 2: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt{147m^7{n}^{11}}=\sqrt{49m^6{n}^{10}·3mn} \\ \sqrt{147m^7{n}^{11}}=\sqrt{49m^6{n}^{10}·3mn} \\ \sqrt{147m^7{n}^{11}}=\sqrt{{\left(7m^3{n}^5\right)}^2}·\sqrt{3mn} \\ \sqrt{147m^7{n}^{11}}=7\left|m^3{n}^5\right|\sqrt{3mn} \end{gathered}$$

Step 3: Using the product property

Using the property, we get

$$\begin{gathered} \sqrt[3]{48x^6{y}^7}=\sqrt[3]{8x^6{y}^6·6y} \\ \sqrt[3]{48x^6{y}^7}=\sqrt[3]{{\left(2x^2{y}^2\right)}^3·6y} \\ \sqrt[3]{48x^6{y}^7}=\sqrt[3]{{\left(2x^2{y}^2\right)}^3}·\sqrt[3]{6y} \\ \sqrt[3]{48x^6{y}^7}=2x^2{y}^2\sqrt[3]{6y} \end{gathered}$$

Step 4: Using the product property

Using the product property, we get

$$\begin{gathered} \sqrt[4]{32x^5{y}^4}=\sqrt[4]{16x^4{y}^4·2x} \\ \sqrt[4]{32x^5{y}^4}=\sqrt[4]{(2xy{)}^4·2x} \\ \sqrt[4]{32x^5{y}^4}=\sqrt[4]{(2xy{)}^4}·\sqrt[4]{2x} \\ \sqrt[4]{32x^5{y}^4}=2\left|xy\right|\sqrt[4]{2x} \end{gathered}$$