Question

Simplify:

$\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}$

Answer 1

Given:

$\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}$

$=\frac{(\sqrt{13}-\sqrt{11}{)}^2+(\sqrt{13}+\sqrt{11}{)}^2}{(\sqrt{13}+\sqrt{11})(\sqrt{13}-\sqrt{11})}$

Since, $(a-b{)}^2=a^2-2ab+b^2$

$(a+b{)}^2=a^2+2ab+b^2$

$(a+b)(a-b)=a^2-b^2$

$\Rightarrow \frac{(\sqrt{13}-\sqrt{11}{)}^2+(\sqrt{13}+\sqrt{11}{)}^2}{(\sqrt{13}+\sqrt{11})(\sqrt{13}-\sqrt{11})}$

$=\frac{13+11-2\sqrt{123}+13+11+2\sqrt{123}}{(\sqrt{13}{)}^2-(\sqrt{11}{)}^2}$

$=\frac{48}{13-11}$

$=\frac{48}{2}$

$=24$

$\therefore \frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}=24$