Question

If $\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=a+b\sqrt{35}$, then the value of $(a-b)$ is :

• A. $7$
• B. $6$
• C. $8$
• D. $Noneofthese$

Answer 1

The correct option is A $7$

$\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$

$=\frac{(\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}$ ............ rationalization of denominator

$=\frac{(\sqrt{7}{)}^2+(\sqrt{5}{)}^2-2\sqrt{7}\sqrt{5}}{(\sqrt{7}{)}^2-(\sqrt{5}{)}^2}$ ............... Using, $(a-b{)}^2=a^2-2ab+b^2$ and $a^2-b^2=(a-b)(a+b)$

$=\frac{12-2\sqrt{35}}{2}$

$⟹\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=6-\sqrt{35}=a+b\sqrt{35}$

So, $a=6,b=-\left(1\right)$

$a-b=6+1=7$