Question

find the value of the above question ​

Answer 1

Given,

$\displaystyle\longrightarrow\sf{x=7+\sqrt{40}}$

$\displaystyle\longrightarrow\sf{x=7+\sqrt{4\times10}}$

$\displaystyle\longrightarrow\sf{x=7+2\sqrt{10}}$

$\displaystyle\longrightarrow\sf{x=5+2+2\sqrt{5\times2}}$

$\displaystyle\longrightarrow\sf{x=\left (\sqrt5\right)^2+\left (\sqrt2\right) ^2+2\cdot\sqrt5\cdot\sqrt2}$

Since $\displaystyle\sf {a^2+b^2+2ab=(a+b)^2,}$

$\displaystyle\longrightarrow\sf{x=\left(\sqrt5+\sqrt2\right)^2}$

$\displaystyle\longrightarrow\sf{\sqrt x=\sqrt5+\sqrt2}$

Then,

$\displaystyle\longrightarrow\sf{\sqrt x+\dfrac {1}{\sqrt x}=\sqrt5+\sqrt2+\dfrac {1}{\sqrt5+\sqrt2}}$

Performing rationalisation,

$\displaystyle\longrightarrow\sf{\sqrt x+\dfrac {1}{\sqrt x}=\sqrt5+\sqrt2+\dfrac {\sqrt5-\sqrt2}{(\sqrt5+\sqrt2)(\sqrt5-\sqrt2)}}$

Since $\displaystyle\sf {(a+b)(a-b)=a^2-b^2,}$

$\displaystyle\longrightarrow\sf{\sqrt x+\dfrac {1}{\sqrt x}=\sqrt5+\sqrt2+\dfrac {\sqrt5-\sqrt2}{5-2}}$

$\displaystyle\longrightarrow\sf{\sqrt x+\dfrac {1}{\sqrt x}=\sqrt5+\sqrt2+\dfrac {\sqrt5-\sqrt2}{3}}$

$\displaystyle\longrightarrow\sf{\sqrt x+\dfrac {1}{\sqrt x}=\dfrac {3\sqrt5+3\sqrt2+\sqrt5-\sqrt2}{3}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {\sqrt x+\dfrac {1}{\sqrt x}=\dfrac {4\sqrt5+2\sqrt2}{3}}}}$