Q. 267 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator. 

$$\begin{gathered} a) \frac{1}{\sqrt[4]{7}} \\ b) \sqrt[4]{\frac{5}{32}} \\ c) \frac{4}{\sqrt[4]{4x^2}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The rationalization of the denominator is $\frac{\sqrt[4]{343}}{7}$.

Part (b) The rationalization of the denominator is $\frac{\sqrt[4]{40}}{4}$.

Part (c) The rationalization of the denominator is $\frac{2\sqrt[4]{4x^2}}{x}$.

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{1}{\sqrt[4]{7}} \\ b) \sqrt[4]{\frac{5}{32}} \\ c) \frac{4}{\sqrt[4]{4x^2}} \end{gathered}$$

Step 2. Solving part (a).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[4]{7^3}$.

$\frac{1}{\sqrt[4]{7}}=\frac{1\times \sqrt[4]{7^3}}{\sqrt[4]{7}\times \sqrt[4]{7^3}}$

Then on simplifying we get,

$\frac{1\times \sqrt[4]{7^3}}{\sqrt[4]{7}\times \sqrt[4]{7^3}}=\frac{\sqrt[4]{343}}{7}$

Hence, the rationalization of the denominator is $\frac{\sqrt[4]{343}}{7}$.

Step 3. Solving part (b).

Now, to rationalize the expression first we will rewrite the expression using quotient property and simplify. 

$$\begin{gathered} \sqrt[4]{\frac{5}{32}}=\frac{\sqrt[4]{5}}{\sqrt[4]{32}} \\ =\frac{\sqrt[4]{5}}{2\sqrt[4]{2}} \end{gathered}$$

Multiplying the denominator and numerator by $\sqrt[4]{2^3}$.

$\frac{\sqrt[4]{5}}{2\sqrt[4]{2}}=\frac{\sqrt[4]{5}\times \sqrt[4]{2^3}}{2\sqrt[4]{2}\times \sqrt[4]{2^3}}$

Then on simplifying we get,

$\frac{\sqrt[4]{5}\times \sqrt[4]{2^3}}{2\sqrt[4]{2}\times \sqrt[4]{2^3}}=\frac{\sqrt[4]{40}}{4}$

Hence, the rationalization of the denominator is $\frac{\sqrt[4]{40}}{4}$.

Step 4. Solving part (c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[4]{4^3{x}^6}$.

$\frac{4}{\sqrt[4]{4x^2}}=\frac{4\times \sqrt[4]{4^3{x}^6}}{\sqrt[4]{4x^2}\times \sqrt[4]{4^3{x}^6}}$

On simplifying we get,

$\frac{4\times \sqrt[4]{4^3{x}^6}}{\sqrt[4]{4x^2}\times \sqrt[4]{4^3{x}^6}}=\frac{8x\sqrt[4]{4x^2}}{4x^2}$

Then removing the common factors.

$\frac{8x\sqrt[4]{4x^2}}{4x^2}=\frac{2\sqrt[4]{4x^2}}{x}$

Hence, the rationalization of the denominator is $\frac{2\sqrt[4]{4x^2}}{x}$.