Question

what is the following quotient?
\frac{2}{\sqrt{13}+\sqrt{11}}
options:
\frac{\sqrt{13}+\sqrt{11}}{6},
\sqrt{13}-\sqrt{11},
\frac{\sqrt{13}+\sqrt{11}}{12},
\sqrt{13}-2\sqrt{11}

Explanation:

Step1: Rationalize the denominator

We have the fraction $\frac{2}{\sqrt{13}+\sqrt{11}}$. To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{13}-\sqrt{11}$. So we get:
$$\frac{2(\sqrt{13}-\sqrt{11})}{(\sqrt{13}+\sqrt{11})(\sqrt{13}-\sqrt{11})}$$

Step2: Apply the difference of squares formula

The denominator is in the form of $(a + b)(a - b)=a^{2}-b^{2}$. Here, $a=\sqrt{13}$ and $b = \sqrt{11}$. So the denominator becomes $(\sqrt{13})^{2}-(\sqrt{11})^{2}=13 - 11=2$. The numerator is $2(\sqrt{13}-\sqrt{11})$. So the fraction simplifies to:
$$\frac{2(\sqrt{13}-\sqrt{11})}{2}$$

Step3: Simplify the fraction

We can cancel out the 2 in the numerator and the denominator. So we are left with $\sqrt{13}-\sqrt{11}$.

Answer:

$\sqrt{13}-\sqrt{11}$ (the option with the text "$\boldsymbol{\sqrt{13}-\sqrt{11}}$")