Question

If $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=a-b\sqrt{6}$, find the values of $a$ and $b$.

• A. $a=2,b=-\frac{5}{4}$
• B. $a=2,b=-\frac{7}{6}$
• C. $a=2,b=-\frac{5}{6}$
• D. $a=2,b=-\frac{3}{6}$

Answer 1

Answer: (C)

$a=2,b=-\frac{5}{6}$

Co-factor of denominator $3\sqrt{2}-2\sqrt{3}$ is $3\sqrt{2}+2\sqrt{3}$.

Multiplying numerator and denominator with co-factor,

$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$

$=\frac{(\sqrt{2}+\sqrt{3})\times (3\sqrt{2}+2\sqrt{3})}{18-12}=\frac{12+5\sqrt{6}}{6}=2-\frac{5}{6}\sqrt{6}$
Comparing this value with $a-b\sqrt{6}$, we get $a=2$ and $b=-\frac{5}{6}$