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$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}+\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=a+b \sqrt{10}$, find a and b.
Given :
The given expression is $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}+\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=a+b \sqrt{10}$.
To do :
We have to find the values of a and b.
Solution :
$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}+\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} = a+b \sqrt{10}$
On cross multiplying,
$\Rightarrow \frac{(\sqrt{5}+\sqrt{2})^2 + (\sqrt{5}-\sqrt{2})^2}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}=a+b \sqrt{10}$
We know that,
$(a+b)^2 = a^2+ b^2 + 2ab$
$(a-b)^2 = a^2+ b^2 - 2ab$
$(a+b)(a-b) = a^2 - b^2$
$\Rightarrow \frac{(\sqrt{5})^2+(\sqrt{2})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{5})^2 +(\sqrt{2})^2 - 2(\sqrt{5})(\sqrt{2}) }{(\sqrt{5})^2 - (\sqrt{2})^2}=a+b \sqrt{10}$
$\Rightarrow \frac{7+2\sqrt{10}+7 - 2\sqrt{10}}{5-2}=a+b \sqrt{10}$
$\Rightarrow \frac{14 + 0\sqrt{10}}{3}=a+b \sqrt{10}$
$\Rightarrow \frac{14}{3} + \frac{0\sqrt{10}}{3}=a+b \sqrt{10}$
On comparing,
$a = \frac{14}{3}$, $b = 0$
The values of a and b are $\frac{14}{3}$ and 0.
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