Q. 260 - Intermediate Algebra

Question

In the following exercises, rationalize the denominator. 

$$\begin{gathered} a) \frac{8}{\sqrt{3}} \\ b) \sqrt{\frac{7}{40}} \\ c) \frac{8}{\sqrt{2y}} \end{gathered}$$

Step-by-Step Solution

Answer

Part(a) The rationalization of the denominator is $\frac{8\sqrt{3}}{3}$.

Part(b) The rationalization of the denominator is $\frac{\sqrt{70}}{20}$.

Part(c) The rationalization of the denominator is $\frac{4\sqrt{2y}}{y}$.

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a) \frac{8}{\sqrt{3}} \\ b) \sqrt{\frac{7}{40}} \\ c) \frac{8}{\sqrt{2y}} \end{gathered}$$

Step 2. Solving part(a).

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

$\frac{8}{\sqrt{3}}=\frac{8\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

On simplifying we get

$\frac{8\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\frac{8\sqrt{3}}{3}$

Hence, the rationalization of the denominator is $\frac{8\sqrt{3}}{3}$.

Step 3. Solving part(b).

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

$$\begin{gathered} \sqrt{\frac{7}{40}}=\frac{\sqrt{7}}{\sqrt{40}} \\ =\frac{\sqrt{7}}{2\sqrt{10}} \end{gathered}$$

Multiplying the numerator and denominator by $\sqrt{10}$.

$\frac{\sqrt{7}}{2\sqrt{10}}=\frac{\sqrt{7}\times \sqrt{10}}{2\sqrt{10}\times \sqrt{10}}$

Then on simplifying we get,

$\frac{\sqrt{7}\times \sqrt{10}}{2\sqrt{10}\times \sqrt{10}}=\frac{\sqrt{70}}{20}$

Hence, the rationalization of the denominator is $\frac{\sqrt{70}}{20}$.

Step 4. Solving part(c).

Now, to rationalize the expression first we will multiply the numerator and denominator by $\sqrt{2y}$.

$\frac{8}{\sqrt{2y}}=\frac{8\times \sqrt{2y}}{\sqrt{2y}\times \sqrt{2y}}$

On simplifying we get,

$\frac{8\times \sqrt{2y}}{\sqrt{2y}\times \sqrt{2y}}=\frac{8\sqrt{2y}}{2y}$

Then removing the common factors.

$\frac{8\sqrt{2y}}{2y}=\frac{4\sqrt{2y}}{y}$

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