Question

Rationalize the numerator.

$\frac{\sqrt{r}+\sqrt{2}}{5}$

Short answer

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

Step-by-step solution

Step 1. Apply the concept of rationalizing the numerator.

If a fraction has a numerator of the form$A+B\sqrt{C}$, we can rationalize the numerator 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

$\frac{\sqrt{r}+\sqrt{2}}{5}$

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

Step 4. Rationalize the numerator.

Consider$\frac{\sqrt{r}+\sqrt{2}}{5}$

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

localid="1645635557620" $\begin{array}{l}\frac{\sqrt{r}+\sqrt{2}}{5}=\frac{\sqrt{r}+\sqrt{2}}{5}\cdot \frac{\sqrt{r}-\sqrt{2}}{\sqrt{r}-\sqrt{2}}\\ \Rightarrow \frac{\sqrt{r}+\sqrt{2}}{5}=\frac{\left(\sqrt{r}+\sqrt{2}\right)\left(\sqrt{r}-\sqrt{2}\right)}{5\left(\sqrt{r}-\sqrt{2}\right)}\\ \Rightarrow \frac{\sqrt{r}+\sqrt{2}}{5}=\frac{{\left(\sqrt{r}\right)}^2-{\left(\sqrt{2}\right)}^2}{5\left(\sqrt{r}-\sqrt{2}\right)}\\ \Rightarrow \frac{\sqrt{r}+\sqrt{2}}{5}=\frac{r-2}{5\left(\sqrt{r}-\sqrt{2}\right)}\end{array}$