Science, asked by ltz02, 2 months ago

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

Answers

Answered by OoINTROVERToO
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}

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

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