Q. 262 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator. 

$$\begin{gathered} a) \frac{4}{\sqrt{5}} \\ b) \sqrt{\frac{27}{80}} \\ c) \frac{18}{\sqrt{6q}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The rationalization of the denominator is $\frac{4\sqrt{5}}{5}$.

Part (b) The rationalization of the denominator is $\frac{\sqrt{135}}{20}$.

Part (c) The rationalization of the denominator is $\frac{3\sqrt{6q}}{q}$.

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{4}{\sqrt{5}} \\ b) \sqrt{\frac{27}{80}} \\ c) \frac{18}{\sqrt{6q}} \end{gathered}$$

Step 2. Solving part (a).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt{5}$.

$\frac{4}{\sqrt{5}}=\frac{4\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$

On simplifying we get,

$\frac{4\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}=\frac{4\sqrt{5}}{5}$

Hence, the rationalization of the denominator is $\frac{4\sqrt{5}}{5}$.

Step 3. Solving part (b).

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

$$\begin{gathered} \sqrt{\frac{27}{80}}=\frac{\sqrt{27}}{\sqrt{80}} \\ =\frac{\sqrt{27}}{4\sqrt{5}} \end{gathered}$$

Multiplying the numerator and denominator by $\sqrt{5}$.

$\frac{\sqrt{27}}{4\sqrt{5}}=\frac{\sqrt{27}\times \sqrt{5}}{4\sqrt{5}\times \sqrt{5}}$

Then on simplifying we get,

$\frac{\sqrt{27}\times \sqrt{5}}{4\sqrt{5}\times \sqrt{5}}=\frac{\sqrt{135}}{20}$

Hence, the rationalization of the denominator is $\frac{\sqrt{135}}{20}$.

Step 4. Solving part (c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt{6q}$.

$\frac{18}{\sqrt{6q}}=\frac{18\times \sqrt{6q}}{\sqrt{6q}\times \sqrt{6q}}$

On simplifying we get,

$\frac{18\times \sqrt{6q}}{\sqrt{6q}\times \sqrt{6q}}=\frac{18\sqrt{6q}}{6q}$

Then removing the common factors.

$\frac{18\sqrt{6q}}{6q}=\frac{3\sqrt{6q}}{q}$

Hence, the rationalization of the denominator is $\frac{3\sqrt{6q}}{q}$.