Q. 268 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator. 

$$\begin{gathered} a) \frac{1}{\sqrt[4]{4}} \\ b) \sqrt[4]{\frac{9}{32}} \\ c) \frac{6}{\sqrt[4]{9x^3}} \end{gathered}$$

Step-by-Step Solution

Answer

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

Part (b) The rationalization of the denominator is $\frac{\sqrt[4]{72}}{4}$.

Part (c) The rationalization of the denominator is $\frac{2\sqrt[4]{9x}}{x}$.

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{1}{\sqrt[4]{4}} \\ b) \sqrt[4]{\frac{9}{32}} \\ c) \frac{6}{\sqrt[4]{9x^3}} \end{gathered}$$

Step 2. Solving part (a).

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

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

On simplifying we get,

$\frac{1\times \sqrt[4]{4^3}}{\sqrt[4]{4}\times \sqrt[4]{4^3}}=\frac{2\sqrt[4]{4}}{4}$

Then removing the common factors.

$\frac{2\sqrt[4]{4}}{4}=\frac{\sqrt[4]{4}}{2}$

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

Step 3. Solving part (b).

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

$$\begin{gathered} \sqrt[4]{\frac{9}{32}}=\frac{\sqrt[4]{9}}{\sqrt[4]{32}} \\ =\frac{\sqrt[4]{9}}{2\sqrt[4]{2}} \end{gathered}$$

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

$\frac{\sqrt[4]{9}}{2\sqrt[4]{2}}=\frac{\sqrt[4]{9}\times \sqrt[4]{2^3}}{2\sqrt[4]{2}\times \sqrt[4]{2^3}}$

Then on simplifying we get,

$\frac{\sqrt[4]{9}\times \sqrt[4]{2^3}}{2\sqrt[4]{2}\times \sqrt[4]{2^3}}=\frac{\sqrt[4]{72}}{4}$

Hence, the rationalization of the denominator is $\frac{\sqrt[4]{72}}{4}$.

Step 4. Solving part (c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[4]{9^3{x}^9}$.

$\frac{6}{\sqrt[4]{9x^3}}=\frac{6\times \sqrt[4]{9^3{x}^9}}{\sqrt[4]{9x^3}\times \sqrt[4]{9^3{x}^9}}$

On simplifying we get,

$\frac{6\times \sqrt[4]{9^3{x}^9}}{\sqrt[4]{9x^3}\times \sqrt[4]{9^3{x}^9}}=\frac{18x^2\sqrt[4]{9x}}{9x^3}$

Then removing the common factors.

$\frac{18x^2\sqrt[4]{9x}}{9x^3}=\frac{2\sqrt[4]{9x}}{x}$

Hence, the rationalization of the denominator is $\frac{2\sqrt[4]{9x}}{x}$.