Question

if a=7+√40 find thevalue of √a+1/√a with solution

Answer 1

We just plug values of a into √a+1/√a then simplify

$\sqrt{a}+\frac{1}{\sqrt{a}}$

$=\sqrt{7+\sqrt{40}}+\frac{1}{\sqrt{7+\sqrt{40}}}$

$=\sqrt{7+\sqrt{40}}*\frac{\sqrt{7+\sqrt{40}}}{\sqrt{7+\sqrt{40}}}+\frac{1}{\sqrt{7+\sqrt{40}}}$

$=\frac{7+\sqrt{40}}{\sqrt{7+\sqrt{40}}}+\frac{1}{\sqrt{7+\sqrt{40}}}$

$=\frac{7+\sqrt{40}+1}{\sqrt{7+\sqrt{40}}}$

$=\frac{8+\sqrt{40}}{\sqrt{7+\sqrt{40}}}$

So the final answer is $=\frac{8+\sqrt{40}}{\sqrt{7+\sqrt{40}}}$.

If you want denominator to be radical free in answer then you can proceed with rationalization method.

Answer 2

$a = 7 + \sqrt{40} = 7 + 2 \sqrt{10} = 5 + 2 + 2 \sqrt{10} = ( \sqrt{5} + \sqrt{2} ) ^{2} \\ \sqrt{a} = \sqrt{5} + \sqrt{2} \\ \frac{1}{ \sqrt{a} } = \frac{1}{ \sqrt{5} + \sqrt{2} } \times \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} - \sqrt{2} } = \frac{ \sqrt{5} - \sqrt{2} }{( \sqrt{5})^{2} -( \sqrt{2}) ^{2} } = \frac{ \sqrt{5} - \sqrt{2} }{5 - 2} = \frac{ \sqrt{5} - \sqrt{2} }{3} \\ now \\ \sqrt{a} + \frac{1}{ \sqrt{a} } = \sqrt{5} + \sqrt{2} + \frac{\sqrt{5} - \sqrt{2}}{3} = \frac{3 \sqrt{5} + 3 \sqrt{2} + \sqrt{5} - \sqrt{2} }{3} = \frac{4 \sqrt{5} + 2 \sqrt{2} }{3}$