Question

Solve the expression: $\frac{3+\sqrt{72}}{3-\sqrt{72}}$

• A. $\frac{9+2\sqrt{2}}{-7}$
• B. $\frac{9+4\sqrt{2}}{-7}$
• C. $\frac{9}{-7}$
• D. $1+\frac{4\sqrt{2}}{9}$

Answer 1

Answer: (B)

$\frac{9+4\sqrt{2}}{-7}$

Given, $\frac{3+\sqrt{72}}{3-\sqrt{72}}$

On rationalizing denominator, we get

$\frac{3+\sqrt{72}}{3-\sqrt{72}}$

$=\frac{(3+\sqrt{72}{)}^2}{9-72}$

$=\frac{9+72+6\sqrt{72}}{-63}$

$=\frac{81+6\sqrt{72}}{-63}$

$=\frac{81+36\sqrt{2}}{-63}$

$=\frac{9+4\sqrt{2}}{-7}$.