Question

sqrt3+sqrt2/sqrt3-sqrt2=a+bsqrt6 find a,b

Answer 1

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

Rationalising the denominator by its conjugate

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

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

$\frac{3+2+2\sqrt{3}.\sqrt{2}}{3-2}<span>=a+b\sqrt{6}</span>$

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

Therefore, a = 5 and b = 2