Question

Find the value of $b$:

$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-b\sqrt{6}$

• A. $b=\frac{-5}{6}$
• B. $b=\frac{5}{6}$
• C. $b=\frac{6}{5}$
• D. $b=-\frac{1}{6}$

Answer 1

The correct option is A $b=\frac{-5}{6}$

Given, $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-b\sqrt{6}$
Multiply and divide it by $3\sqrt{2}+2\sqrt{3}$
$\frac{(\sqrt{2}+\sqrt{3})(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})}$ = $2-b\sqrt{6}$
$\frac{6+3\sqrt{6}+2\sqrt{6}+6}{6}$ = $2-b\sqrt{6}$
$\frac{12+5\sqrt{6}}{6}=2-b\sqrt{6}$
Hence, $b=\frac{-5}{6}$