Question

if x=3+2√2 find the value of (√x-1/√x)​

Answer 1

Answer :

$x = 3 + 2 \sqrt{2} \\ \\ x = 2 + 1 + 2 \sqrt{2} \\ \\x = ( \sqrt{2} ) {}^{2} + 1 {}^{2} + 2 \times \sqrt{2 } \times 1 \\ \\ x = ( \sqrt{2} + 1 ) {}^{2} \\ \\ \sqrt{x} = \sqrt{( \sqrt{2} + 1) {}^{2} } \\ \\ \sqrt{x} = \sqrt{2} + 1$

Now,

$\frac{ 1}{ \sqrt{x}} = \frac{1}{ \sqrt{2} + 1} \\ \\ \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1}{ \sqrt{2} - 1} \\ \\ \frac{1}{ \sqrt{x} } = \frac{ \sqrt{2} - 1 }{( \sqrt{2}) {}^{2} - 1 {}^{2} } \\ \\ \frac{1}{\sqrt{x} } = \sqrt{2} - 1$

Now

$\sqrt{x} - \frac{1}{ \sqrt{x} } = \sqrt{2 }+ 1 - (\sqrt{2} - 1) \\ \\ \sqrt{x} - \frac{1}{ \sqrt{x} } = \sqrt{2 } + 1 - \sqrt{2} + 1 \\ \\ \sqrt{x} - \frac{1}{ \sqrt{x} } = 1 + 1 \\ \\ \boxed{ \bf \sqrt{x} - \frac{1}{ \sqrt{x} } = 2}$

Answer 2

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