Math, asked by anjalisingh2288, 14 days ago

yar please answer me wiyh solution ​

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Answers

Answered by BrainlyArnab
1

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.

Answered by abhradeepde
2

Step-by-step explanation:

n = 3 - 2 \sqrt{2 }  \\  =  >n =   {( \sqrt{2} )}^{2}  + 1 - 2 \sqrt{2}  \\  =  > n =  {( \sqrt{2 } - 1) }^{2}  \\   =  >  \sqrt{n}  =  \sqrt{2}  - 1 \\  \\ so \:  \frac{1}{ \sqrt{n} }  =  \frac{1}{ \sqrt{2}  - 1}  \\  =  >  \frac{1}{n}  =  \sqrt{2}  + 1 \\  \\ now \\  \sqrt{n}   -   \frac{1}{ \sqrt{n} }  =  \sqrt{2}  - 1  -  \sqrt{2}   - 1 \\  =  - 2

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