Question

Question 14
After rationalizing the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$, we get the denominator as

A) 13
B) 19
C) 5
D) 35

Answer 1

The answer is B.
$\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{7}{3\sqrt{3}-2\sqrt{2}}.\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$
[ Multiplying numerator and denominator by $3\sqrt{3}+2\sqrt{2}]$
[ using identity $(a-b)(a+b)=a^2-b^2$]
$\frac{7(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3}{)}^2-(2\sqrt{2}{)}^2}=\frac{7(3\sqrt{3}+2\sqrt{2})}{27-8}$
$[\because \text{fraction}=\frac{\text{numerator}}{\text{denominator}}]=\frac{7(3\sqrt{3}+2\sqrt{2})}{19}$
Hence, after rationalizing the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$ , we get the denominator as 19.