Question

If ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ are rational numbers, find the value of $a$ and $b$ in each of the following equations.
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=a+b\sqrt{6}$

Answer 1

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

consider,$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

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

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

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

=$5+2\sqrt{6}$

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

Here a=5 and b=2