Question

yar please answer me​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$x = 3 - 2 \sqrt{2} \\ ( \sqrt{x} - \frac{1}{ \sqrt{x} }) = ± 2 \\ \\ \sqrt{x} - \frac{1}{ \sqrt{x} } = ± 2 \\ \\ = > ( \sqrt{x} - \frac{1}{ \sqrt{x} } {)}^{2} = ( ± 2 {)}^{2} \\ (by \: squaring \: both \: sides) \\ \\ = > {( \sqrt{x} ) }^{2} - 2( \sqrt{x} )( \frac{1}{ \sqrt{x} } ) + ( \frac{1}{ \sqrt{x} } {)}^{2} = 4 \\ (using \: formula {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} ) \\ \\ = > x - 2 + \frac{1}{x} = 4 \\ \\ = > x + \frac{1}{x} = 4 + 2 \\ \\ = > \frac{ {x}^{2} + 1}{x} = 6 \\ \\ = > {x}^{2} + 1 = 6x \\ \\ = > {x}^{2} - 6x + 1 = 0$

Now it is a quadratic equation (ax² + bx + c = 0)

a = 1, b = -6, c = 1

Using the quadratic equation formula,

$\frac{ - b ± \sqrt{ {b}^{2 } - 4ac} }{2a} \\ \\ = > \frac{ - ( - 6) ± \sqrt{ {( - 6)}^{2} - 4(1)(1) } }{2(1)} \\ \\ = > \frac{6 ± \sqrt{36 - 4} }{2} \\ \\ = > \frac{6 ± \sqrt{32} }{2} \\ \\ = > \frac{6 ± \sqrt{16 \times 2} }{2} \\ \\ = > \frac{6 ± 4 \sqrt{2} }{2} \\ \\ = > 3 ± 2 \sqrt{2} \\ \\ = > x = 3 + 2 \sqrt{2} \: and \: 3 - 2 \sqrt{2} \\ \\ \huge \sf \: hence \: proved \: that \: x = 3 - 2 \sqrt{2}$

hope it helps.