Question

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

• A. $(5,2)$
• B. $(2,5)$
• C. $(3,2)$
• D. $(1,5)$

Answer 1

The correct option is A $(5,2)$

$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ Rationalising numerator and denominator .

$\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\frac{3+2+2\sqrt{6}}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}=\frac{5+2\sqrt{6}}{1}$

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

$a=5;b=2$

$(a,b)=(5,2)$