Question

Rationalize the denominator.

$\frac{y}{\sqrt{3}+\sqrt{y}}$

Short answer

The solution is$\frac{y\left(\sqrt{3}-\sqrt{y}\right)}{3-y}$

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

$\frac{y}{\sqrt{3}+\sqrt{y}}$

Here the denominator is $\sqrt{3}+\sqrt{y}$, so multiply the numerator and denominator by$\sqrt{3}-\sqrt{y}$

Step 4. Rationalize the denominator.

Consider$\frac{y}{\sqrt{3}+\sqrt{y}}$

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

role="math" localid="1643266022516" $\begin{array}{l}\frac{y}{\sqrt{3}+\sqrt{y}}=\frac{y}{\sqrt{3}+\sqrt{y}}\cdot \frac{\sqrt{3}-\sqrt{y}}{\sqrt{3}-\sqrt{y}}\\ \Rightarrow \frac{y}{\sqrt{3}+\sqrt{y}}=\frac{y\left(\sqrt{3}-\sqrt{y}\right)}{\left(\sqrt{3}+\sqrt{y}\right)\left(\sqrt{3}-\sqrt{y}\right)}\\ \Rightarrow \frac{y}{\sqrt{3}+\sqrt{y}}=\frac{y\left(\sqrt{3}-\sqrt{y}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{y}\right)}^2}\\ \Rightarrow \frac{y}{\sqrt{3}+\sqrt{y}}=\frac{y\left(\sqrt{3}-\sqrt{y}\right)}{3-y}\end{array}$