Question

$\displaystyle\sf x = 3+2\sqrt{2}$
$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$??
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Answer 1

Answer

\displaystyle\sf x = 3+2\sqrt{2}x=3+22

\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}x−x1

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\begin{gathered}\displaystyle\sf :\implies \dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}}\\\end{gathered}:⟹x1=3+221

\displaystyle\sf :\implies \dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}:⟹3+221×3−223−22

\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{3^2-(2\sqrt{2}^2)}:⟹32−(222)3−22

\displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{9-8}:⟹9−83−22

\displaystyle\sf :\implies \dfrac{1}{x} = 3-2\sqrt{2}:⟹x1=3−22

__________________

\displaystyle\sf :\implies x+\dfrac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2}):⟹x+x1=(3+22)+(3−22)

\displaystyle\sf :\implies x+\dfrac{1}{x} = 3+2\sqrt{2} + 3 - 2\sqrt{2}:⟹x+x1=3+22+3−22

\displaystyle\sf :\implies x+\dfrac{1}{x} = 6:⟹x+x1=6

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} = 2+4:⟹x+x1=2+4

\displaystyle\sf :\implies x+\dfrac{1}{x}-2 = 4:⟹x+x1−2=4

\displaystyle\sf :\implies \bigg\lgroup \sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg\rgroup^2 = 4:⟹⎩⎪⎪⎪⎧x−x1⎭⎪⎪⎪⎫2=4

\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \sqrt{4}:⟹x−x1=4

\displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \pm 2:⟹x−x1=±2

\displaystyle\therefore\:\underline{\textsf{The value of $ \sqrt{ \sf x}-\dfrac{\sf 1}{\sqrt{\sf x}}$ is \textbf{$\pm$2 }}}∴The value of x−x1 is ±2

Hope it will help you!!

Answer 2

Explanation:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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