Question

If $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{7+b\sqrt{10}}{a}$. Then $(a+b)$ is

Answer 1

Given:

$\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{7+b\sqrt{1}0}{a}$

$RationaliseLHS$

$\Rightarrow \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{7+b\sqrt{1}0}{a}$

$=\frac{(\sqrt{5}-\sqrt{2}{)}^2}{(\sqrt{5}+\sqrt{2})\times (\sqrt{5}-\sqrt{2})}=\frac{7+b\sqrt{1}0}{a}$

$=\frac{5+2-2\sqrt{1}0}{(\sqrt{5}{)}^2+(\sqrt{2}{)}^2}=\frac{7+b\sqrt{1}0}{a}$

$\Rightarrow \frac{7+(-2\sqrt{1}0)}{3}=\frac{7+b\sqrt{1}0}{a}$

Compare respective term of LHS and RHS

$a=3andb=-2$

Hence,

$a+b=1$