Question

$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}$ is equal to

• A. $1$
• B. $2$
• C. $6-\sqrt{35}$
• D. $6+\sqrt{35}$

Answer 1

The correct option is D $6+\sqrt{35}$

To find the value of $\left(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}\right)$

Rationalising the denominator

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

$=\frac{(\sqrt{7}+\sqrt{5}{)}^2}{7-5}$

$=\frac{7+5+2\times \sqrt{7}\times \sqrt{5}}{2}$

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

$=6+\sqrt{35}$

Hence, the required value is $6+\sqrt{35}$ .