Question

If $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=a-b\sqrt{6}$, find rational numbers a and b.

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

Answer 1

Answer: (C)

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

Given $\left(\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}\right)$

$=\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{6}(\sqrt{3}-\sqrt{2})}\right)$

Rationalize the denominator

$=\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{6}(\sqrt{3}-\sqrt{2})}\right)\times \left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right)$

$=\left(\frac{2+3+2\sqrt{6}}{\sqrt{6}\times (3-2)}\right)$

$=\left(\frac{5}{\sqrt{6}}\right)+2$

$=2+\left(\frac{5}{6}\right)\sqrt{6}$

by comparing

$\therefore a=2andb=\left(\frac{-5}{6}\right)$