Question

$\huge \displaystyle\sf x = 3+2\sqrt{2}$


$\huge \displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}} = ¿?$ ​

Answer 1

$\displaystyle\sf x = 3+2\sqrt{2}$

$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$

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$\begin{gathered}\displaystyle\sf :\implies \dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}}\\\end{gathered}$

$\displaystyle\sf :\implies \dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{3^2-(2\sqrt{2}^2)}$

$\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{9-8}$

$\displaystyle\sf :\implies \dfrac{1}{x} = 3-2\sqrt{2}$

__________________

$\displaystyle\sf :\implies x+\dfrac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2})$

$\displaystyle\sf :\implies x+\dfrac{1}{x} = 3+2\sqrt{2} + 3 - 2\sqrt{2}$

$\displaystyle\sf :\implies x+\dfrac{1}{x}$

So here we know that we may split the number 6 into 4+2 and 4+2 = 6

$\displaystyle\sf :\implies x+\dfrac{1}{x} = 4$

$\displaystyle\sf :\implies x+\dfrac{1}{x}-2 = 4$

$\displaystyle\sf :\implies \bigg\lgroup \sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg\rgroup^2 = 4$

$\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \sqrt{4}$

$\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \pm 2$

$\displaystyle\therefore\:\underline{\textsf{The value of \sqrt{ \sf x}-\dfrac{\sf 1}{\sqrt{\sf x}} is \textbf{ \pm 2 }}}$

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Answer 2

Answer:

Answer:

$\huge\bf\underline \red{\underline{Answer}}$

$\displaystyle\sf x = 3+2\sqrt{2}$

$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$

━━━━━━━━

$\begin{gathered}\displaystyle\sf :\implies \dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}}\\\end{gathered}$

$\displaystyle\sf :\implies \dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{3^2-(2\sqrt{2}^2)}$

$\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{9-8}$

$\displaystyle\sf :\implies \dfrac{1}{x} = 3-2\sqrt{2}$

__________________

$\displaystyle\sf :\implies x+\dfrac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2})$

$\displaystyle\sf :\implies x+\dfrac{1}{x} = 3+2\sqrt{2} + 3 - 2\sqrt{2}$

$\displaystyle\sf :\implies x+\dfrac{1}{x}$

So here we know that we may split the number 6 into 4+2 and 4+2 = 6

$\displaystyle\sf :\implies x+\dfrac{1}{x} = 4$

$\displaystyle\sf :\implies x+\dfrac{1}{x}-2 = 4$

$\displaystyle\sf :\implies \bigg\lgroup \sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg\rgroup^2 = 4$

$\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \sqrt{4}$

$\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \pm 2$

$\displaystyle\therefore\:\boxed{\textsf{The value of \sqrt{ \sf x}-\dfrac{\sf 1}{\sqrt{\sf x}} is \textbf{ \pm 2 }}}$

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