Q. 270 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator. 

$$\begin{gathered} a) \frac{1}{\sqrt[4]{8}} \\ b) \sqrt[4]{\frac{27}{128}} \\ c) \frac{16}{\sqrt[4]{64b^2}} \end{gathered}$$

Step-by-Step Solution

Answer

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

Part (b) The rationalization of the denominator is $\frac{\sqrt[4]{54}}{4}$.

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

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{1}{\sqrt[4]{8}} \\ b) \sqrt[4]{\frac{27}{128}} \\ c) \frac{16}{\sqrt[4]{64b^2}} \end{gathered}$$

Step 2. Solving part (a).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[4]{8^3}$.

$\frac{1}{\sqrt[4]{8}}=\frac{1\times \sqrt[4]{8^3}}{\sqrt[4]{8}\times \sqrt[4]{8^3}}$

On simplifying we get,

$\frac{1\times \sqrt[4]{8^3}}{\sqrt[4]{8}\times \sqrt[4]{8^3}}=\frac{4\sqrt[4]{2}}{8}$

Then removing the common factors.

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

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

Step 3. Solving Part (b).

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

$$\begin{gathered} \sqrt[4]{\frac{27}{128}}=\frac{\sqrt[4]{27}}{\sqrt[4]{128}} \\ =\frac{\sqrt[4]{27}}{2\sqrt[4]{8}} \end{gathered}$$

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

$\frac{\sqrt[4]{27}}{2\sqrt[4]{8}}=\frac{\sqrt[4]{27}\times \sqrt[4]{2}}{2\sqrt[4]{8}\times \sqrt[4]{2}}$

On simplifying we get,

$\frac{\sqrt[4]{27}\times \sqrt[4]{2}}{2\sqrt[4]{8}\times \sqrt[4]{2}}=\frac{\sqrt[4]{54}}{4}$

Hence, the rationalization of the denominator is $\frac{\sqrt[4]{54}}{4}$.

Step 4. Solving part (c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt[4]{{\left(64\right)}^3{b}^6}$.

$\frac{16}{\sqrt[4]{64b^2}}=\frac{16\times \sqrt[4]{{\left(64\right)}^3{b}^6}}{\sqrt[4]{64b^2}\times \sqrt[4]{{\left(64\right)}^3{b}^6}}$

On simplifying we get,

$\frac{16\times \sqrt[4]{{\left(64\right)}^3{b}^6}}{\sqrt[4]{64b^2}\times \sqrt[4]{{\left(64\right)}^3{b}^6}}=\frac{256b\sqrt[4]{4b^2}}{64b^2}$

Then removing the common factors.

$\frac{256b\sqrt[4]{4b^2}}{64b^2}=\frac{4\sqrt[4]{4b^2}}{b}$

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