Question

If $a$ and $b$ 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

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

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

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

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

$a=5$ & $b=2$