Question:
After rationalising the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}}$, we get the denominator as
(a) 13
(b) 19
(c) 5
(d) 35
Solution:
$\frac{7}{3 \sqrt{3}-2 \sqrt{2}}=\frac{7}{3 \sqrt{3}-2 \sqrt{2}} \cdot \frac{3 \sqrt{3}+2 \sqrt{2}}{3 \sqrt{3}+2 \sqrt{2}}$ [multiplying numerator and denominator by $3 \sqrt{3}+2 \sqrt{2}$ ]
$=\frac{7(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}}$
[using identity $(a-b)(a+b)=a^{2}-b^{2}$ ]
$=\frac{7(3 \sqrt{3}+2 \sqrt{2})}{27-8}$ $\left[\because\right.$ fraction $\left.=\frac{\text { numerator }}{\text { denominator }}\right]$
$=\frac{7(3 \sqrt{3}+2 \sqrt{2})}{19}$
Hence, after rationalising the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}}$, we get the denominator as 19 .