Question

Rationalize the denominator.

$\frac{2}{\sqrt{2}+\sqrt{7}}$

Short answer

The solution is$-\frac{2}{5}\left(\sqrt{2}-\sqrt{7}\right)$

Step-by-step solution

Step 1. Apply the concept of rationalizing the denominator.

If a fraction has a denominator of the form$A+B\sqrt{C}$, we can rationalize the denominator by multiplying numerator and denominator by the conjugate radical $A-B\sqrt{C}$.

Step 2. Apply the concept of special multiplication.

If A and B are real numbers, then

$\left(A-B\right)\left(A+B\right)=A^2-B^2$

Step 3. Analyze the expression.

Given

$\frac{2}{\sqrt{2}+\sqrt{7}}$

Here the denominator is $\sqrt{2}+\sqrt{7}$, so multiply the numerator and denominator by$\sqrt{2}-\sqrt{7}$

Step 4. Rationalize the denominator.

Consider$\frac{2}{\sqrt{2}+\sqrt{7}}$

So, the simplified expression after rationalizing denominator is given by:

role="math" localid="1643226641463" $\begin{array}{l}\frac{2}{\sqrt{2}+\sqrt{7}}=\frac{2}{\sqrt{2}+\sqrt{7}}\cdot \frac{\sqrt{2}-\sqrt{7}}{\sqrt{2}-\sqrt{7}}\\ \Rightarrow \frac{2}{\sqrt{2}+\sqrt{7}}=\frac{2\left(\sqrt{2}-\sqrt{7}\right)}{\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)}\\ \Rightarrow \frac{2}{\sqrt{2}+\sqrt{7}}=\frac{2\left(\sqrt{2}-\sqrt{7}\right)}{{\left(\sqrt{2}\right)}^2-{\left(\sqrt{7}\right)}^2}\\ \Rightarrow \frac{2}{\sqrt{2}+\sqrt{7}}=\frac{2\left(\sqrt{2}-\sqrt{7}\right)}{2-7}\\ \Rightarrow \frac{2}{\sqrt{2}+\sqrt{7}}=\frac{2\left(\sqrt{2}-\sqrt{7}\right)}{-5}\\ \Rightarrow \frac{2}{\sqrt{2}+\sqrt{7}}=-\frac{2}{5}\left(\sqrt{2}-\sqrt{7}\right)\end{array}$