Question

Find the value of ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$, if : $\frac{2\sqrt{3}+3\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}=a+b\sqrt{6}$

Answer 1

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

Rationalising the L.H.S:

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

$=\frac{{(2\sqrt{3}+3\sqrt{2})}^2}{(2\sqrt{3}{)}^2-(3\sqrt{2}{)}^2}$

$=\frac{(2\sqrt{3}{)}^2+(3\sqrt{2}{)}^2+2\times 2\sqrt{3}\times 3\sqrt{2}}{12-18}$

$=\frac{12+18+12\sqrt{6}}{-6}$

$=\frac{30+12\sqrt{6}}{-6}$

$=-5-2\sqrt{6}$

Now,

$-5-2\sqrt{6}=a+b\sqrt{6}$

Therefore, $a=-5$ and $b=-2$