Question

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

Answer 1

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...
☺☺

Answer 2

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