Q. 8.34 - Intermediate Algebra

Question

Simplify:

$\left(a\right) \sqrt{180m^9{n}^{11}} \left(b\right) \sqrt{72x^6{y}^5} \left(c\right) \sqrt{80x^7{y}^4}.$

Step-by-Step Solution

Answer

Part a. After simplifying we get, $6m^4{n}^5\sqrt{5mn}.$

Part b. After simplifying we get, $2x^{2}y \sqrt[3]{9y^2}.$

Part c. After simplifying we get, $2\left|xy\right|\sqrt[4]{5x^3}.$

Part a Step 1. Given Information

The given radical is $\sqrt{180m^9{n}^{11}}$.

We have to simplify it.

Part a Step 2. Rewrite the radical as the product of three radicals

So, we can write the radical as

$\sqrt{180}\sqrt{m^9}\sqrt{n^{11}}$

Part a Step 3. Take the square root

By taking the square root we get,

$$\begin{gathered} =\sqrt{180}\sqrt{m^9}\sqrt{n^{11}} \\ =\sqrt{36m^8{n}^{10}}\sqrt{5mn} \\ =6m^4{n}^5\sqrt{5mn} \end{gathered}$$

By simplifying we get $6m^4{n}^5\sqrt{5mn}.$

Part b Step 1. Given Information

The given radical is $\sqrt[3]{72x^6{y}^5}.$

We have to simplify it.

Part b Step 2. Rewrite the radicand as a product

To rewrite the radicand as a product we will be using the largest perfect cube factor.

So, $\sqrt[3]{\left(2^3{x}^6{y}^3\right)\times \left(9y^2\right)}$

Part b Step 3. Take the cube root

By taking the cube root we get,

$$\begin{gathered} =\sqrt[3]{\left(2^3{x}^6{y}^3\right)\times \left(9y^2\right)} \\ =2x^{2}y \sqrt[3]{9y^2} \end{gathered}$$

By simplifying we get $2x^{2}y\sqrt[3]{9y^2}$

Part c Step 1. Given Information

The given radical is $\sqrt[4]{80x^7{y}^4}.$

We have to simplify it.

Part c Step 2. Rewrite the radicand as a product

To rewrite the radicand as a product we will be using the largest perfect fourth power factor.

So, $$\begin{gathered} \sqrt[4]{80x^7{y}^4} \\ =\sqrt[4]{\left(2^4{x}^4{y}^4\right)\times \left(5x^3\right)} \end{gathered}$$ 

Part c Step 3. Simplify

By simplifying we get,

$$\begin{gathered} =\sqrt[4]{\left(2^4{x}^4{y}^4\right)\times \left(5x^3\right)} \\ =2\left|xy\right|\sqrt[4]{5x^3} \end{gathered}$$