Q. 269 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator. 

$$\begin{gathered} a) \frac{1}{\sqrt[4]{9}} \\ b) \sqrt[4]{\frac{25}{128}} \\ c) \frac{6}{\sqrt[4]{27a}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The rationalization of the denominator is $\frac{\sqrt[4]{9}}{3}$.

Part (b) The rationalization of the denominator is $\frac{\sqrt[4]{50}}{4}$.

Part (c) The rationalization of the denominator is $\frac{2\sqrt[4]{3a^3}}{a}$.

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{1}{\sqrt[4]{9}} \\ b) \sqrt[4]{\frac{25}{128}} \\ c) \frac{6}{\sqrt[4]{27a}} \end{gathered}$$

Step 2. Solving part (a).

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

$\frac{1}{\sqrt[4]{9}}=\frac{1\times \sqrt[4]{9^3}}{\sqrt[4]{9}\times \sqrt[4]{9^3}}$

On simplifying we get,

$\frac{1\times \sqrt[4]{9^3}}{\sqrt[4]{9}\times \sqrt[4]{9^3}}=\frac{3\sqrt[4]{9}}{9}$

Then removing the common factors.

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

Hence, the rationalization of the denominator is $\frac{\sqrt[4]{9}}{3}$.

Step 3 Solving part (c).

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

$$\begin{gathered} \sqrt[4]{\frac{25}{128}}=\frac{\sqrt[4]{25}}{\sqrt[4]{128}} \\ =\frac{\sqrt[4]{25}}{2\sqrt[4]{8}} \end{gathered}$$

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

$\frac{\sqrt[4]{25}}{2\sqrt[4]{8}}=\frac{\sqrt[4]{25}\times \sqrt[4]{2}}{2\sqrt[4]{8}\times \sqrt[4]{2}}$

Then on simplifying we get,

$\frac{\sqrt[4]{25}\times \sqrt[4]{2}}{2\sqrt[4]{8}\times \sqrt[4]{2}}=\frac{\sqrt[4]{50}}{4}$

Hence, the rationalization of the denominator is $\frac{\sqrt[4]{50}}{4}$.

Step 4. Solving Part (c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[4]{{\left(27\right)}^3{a}^3}$.

$\frac{6}{\sqrt[4]{27a}}=\frac{6\times \sqrt[4]{{\left(27\right)}^3{a}^3}}{\sqrt[4]{27a}\times \sqrt[4]{{\left(27\right)}^3{a}^3}}$

Then on simplifying we get,

$\frac{6\times \sqrt[4]{{\left(27\right)}^3{a}^3}}{\sqrt[4]{27a}\times \sqrt[4]{{\left(27\right)}^3{a}^3}}=\frac{54\sqrt[4]{3a^3}}{27a}$

Finally, removing the common factors.

$\frac{54\sqrt[4]{3a^3}}{27a}=\frac{2\sqrt[4]{3a^3}}{a}$

Hence, the rationalization of the denominator is $\frac{2\sqrt[4]{3a^3}}{a}$.