Math, asked by kalpit2, 1 year ago

if x = to 2 + root 3 find X square + 1 upon x square

Answers

Answered by DaIncredible
2
Hey friend,
Here is the answer you were looking for:
x = 2 +  \sqrt{3}  \\  \\  \frac{1}{x}  =  \frac{1}{2 +  \sqrt{3} }

On rationalizing the denominator we get,

 \frac{1}{x}  =  \frac{1}{2 +  \sqrt{3} }  \times  \frac{2 -  \sqrt{3} }{2 -  \sqrt{3} }  \\

Using the identity :

( x + y)(x - y) =  {x}^{2}  -  {y}^{2}

 \frac{1}{x}  =  \frac{2 -  \sqrt{3} }{ {(2)}^{2}  -  {( \sqrt{3} })^{2} }  \\  \\   \frac{1}{x}  =  \frac{2 -  \sqrt{3} }{4 - 3}  \\  \\  \frac{1}{x}  = 2 -  \sqrt{3}  \\  \\ x +  \frac{1}{x}  = (2  +  \sqrt{3} ) + (2 -  \sqrt{3} ) \\  \\ x +  \frac{1}{x}  = 2 +  \sqrt{3}  + 2 -  \sqrt{3}  \\  \\ x +  \frac{1}{x}  =   4 \\

Squaring both the sides we get :

 {(x +  \frac{1}{x} })^{2}  =  {(4)}^{2}  \\

Using the identity :

 {(x + y)}^{2}  =  {x}^{2}  +  {y}^{2} + 2xy
 {x}^{2}  +  {( \frac{1}{x} )}^{2}  + 2 \times x \times  \frac{1}{x}  = 16 \\  \\  {x}^{2}  +  \frac{1}{ {x}^{2} }  + 2 = 16 \\  \\  {x}^{2}  +  \frac{1}{ {x}^{2} }  = 16 - 2 \\  \\  {x}^{2}  +  \frac{1}{ {x}^{2} }  = 14

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
☺☺
Answered by SmãrtyMohït
11
Here is your solution

Given :-

x=2+√3

Now

 \frac{1}{x} = \frac{1}{2 +\sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{2 {}^{2} - ( \sqrt{3}) {}^{2} } \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} \\ \frac{1}{x} = 2 - \sqrt{3} \\ \\ \\

 x + \frac{1}{x} = 2 - \sqrt{3} + 2 + \sqrt{3} \\ x + \frac{1}{x} = 4 \\ Both \: sides \: squaring. \: \\ (x + \frac{1}{x} ) {}^{2} = 4 {}^{2} \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 16 \\ x {}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 14

Hope it helps you
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