Q. 8.33 - Intermediate Algebra

Question

Simplify-

(a) $\sqrt{98a^7{b}^5}$

(b) $\sqrt[3]{56x^5{y}^4}$

(c) $\sqrt[4]{32x^5{y}^8}$

Step-by-Step Solution

Answer

The values are-

(a) $7b^2\left|a^3\right|\sqrt{2ab}$

(b) $2xy\sqrt[3]{7x^{2}y}$

(c) $2y^2\left|x\right|\sqrt[4]{2x}$

Step 1. Explanation

(a) We will write the radical in form of product of two radicals with one being the largest perfect square factor of the number. Then we will simplify it to get the value. 

(b)  We will write the radical in form of product of two radicals with one being the largest perfect cube factor of the number. Then we will simplify it to get the value.  

(c) We will write the radical in form of product of two radicals with one being the largest perfect fourth power factor of the number. Then we will simplify it to get the value.

Step 2. Solving for part (a)

We have $\sqrt{98a^7{b}^5}$.

We have $49a^6{b}^4$ as the greatest perfect square factor of $98a^7{b}^5$.

So this can be written as $\sqrt{49a^6{b}^4\times 2ab}$.

Using the product rule, we get
$$\begin{gathered} =\sqrt{49a^6{b}^4}\times \sqrt{2ab} \\ =\sqrt{{\left(7a^3{b}^2\right)}^2}\times \sqrt{2ab} \\ =7b^2\left|a^3\right|\sqrt{2ab} \end{gathered}$$

Thus we get the value as $7b^2\left|a^3\right|\sqrt{2ab}$.

Step 3. Solving for part (b)

We have $\sqrt[3]{56x^5{y}^4}$.

We have $8x^3{y}^3$ as the greatest perfect cube factor of $56x^5{y}^4$.

So this can be written as $\sqrt[3]{8x^3{y}^3\times 7x^{2}y}$.

Using the product rule, we get

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

Thus we get the value as $2xy \sqrt[3]{7x^{2}y}$.

Step 4. Solving for part (c)

We have $\sqrt[4]{32x^5{y}^8}$.

We have $16x^4{y}^8$ as the greatest perfect fourth power factor of $32x^5{y}^8$.

So this can be written as $\sqrt[4]{16x^4{y}^8\times 2x}$.

Using the product rule, we get
$$\begin{gathered} =\sqrt[4]{16x^4{y}^8}\times \sqrt[4]{2x} \\ =\sqrt[4]{{\left(2x{y}^2\right)}^4}\times \sqrt[4]{2x} \\ =2y^2\left|x\right| \sqrt[4]{2x} \end{gathered}$$

Thus we get the value as $2y^2\left|x\right| \sqrt[4]{2x}$.