Q. 8.104 - Intermediate Algebra

Question

In the following exercise ,simplify 

$$\begin{gathered} \left(a\right) \frac{1}{\sqrt[4]{5}} \\ \left(b\right) \sqrt[4]{\frac{7}{128}} \\ \left(c\right) \frac{4}{\sqrt[4]{4x}} \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right) \frac{\sqrt[4]{5^3}}{5} \\ \left(b\right) \frac{\sqrt[4]{56}}{4} \\ \left(c\right) \frac{\sqrt[4]{4^3{x}^3}}{x} \end{gathered}$$ 

Part (a) Step 1. Given information.

Given expression:

$\frac{1}{\sqrt[4]{5}}$

Part (a) Step 2. Finding solution.

Finding the solution.

Multiply numerator and denominator by the same factor.

Multiply numerator and denominator $\sqrt[4]{5^3}$.

$$\begin{gathered} \frac{1}{\sqrt[4]{5}}\times \frac{\sqrt[4]{5^3}}{\sqrt[4]{5^3}} \\ =\frac{\sqrt[4]{5^3}}{\sqrt[4]{5^{3+1}}} \\ =\frac{\sqrt[4]{5^3}}{5} \end{gathered}$$

Part (b) Step 1. Given information.

Given expression:

$\sqrt[4]{\frac{7}{128}}$

Part (b) Step 2. Finding the solution.

Finding the solution.

Multiply numerator and denominator by the same factor.

Multiply numerator and denominator $\sqrt[4]{2^3}$.

$$\begin{gathered} \frac{\sqrt[4]{7}}{\sqrt[4]{128}}=\frac{\sqrt[4]{7}}{\sqrt[4]{2^7}} \\ =\frac{\sqrt[4]{7}}{\sqrt[4]{2^4.2^3}} \\ =\frac{\sqrt[4]{7}}{\sqrt[4]{2^4}.\sqrt[4]{2^3}} \\ =\frac{\sqrt[4]{7}}{2.\sqrt[4]{2^3}} \\ =\frac{\sqrt[4]{7}}{2.\sqrt[4]{2^3}}\times \frac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}} \\ =\frac{\sqrt[4]{7}.\sqrt[4]{2^3}}{2.\sqrt[4]{2^{3+1}}} \\ =\frac{\sqrt[4]{7}.\sqrt[4]{2^3}}{2.2} \\ =\frac{\sqrt[4]{56}}{4} \end{gathered}$$

Part (c) Step 1. Given information.

Given expression :

$\frac{4}{\sqrt[4]{4x}}$

Part (c) Step 2. Finding the solution.

Finding the solution.

Multiply numerator and denominator by the same factor.

Multiply numerator and denominator $\sqrt[4]{4^3{x}^3}$.

$$\begin{gathered} \frac{4}{\sqrt[4]{4x}}\times \frac{\sqrt[4]{4^3{x}^3}}{\sqrt[4]{4^3{x}^3}} \\ =\frac{4\sqrt[4]{4^3{x}^3}}{4x} \\ =\frac{\sqrt[4]{4^3{x}^3}}{x} \end{gathered}$$