Question

if x is equal to 2 + root 3 find the value of a square + 1 x square

Answer 1

Hi there!

Given :-
x = 2 + √3

So, x² = (2 + √3)² = 4 + 3 + 4√3 = 7 + 4√3

Now,

1 / x = 1 / 2 + √3

= 1 / 2 + √3 × 2 - √3 / 2 - √3

= 2 - √3 / 4 - 3

= 2 - √3

Thus,

(1 / x)² = (2 - √3)²

= 4 + 3 - 4√3

= 7 - 4√3

Therefore,

x² + (1 / x)² = 7 + 4√3 + 7 - 4√3

= 7 + 7

= 14

Hence, The required answer is :-
Value of x² + (1 / x)² = 14

hope it helps! :)

Answer 2

Heya bro,
I guess you have written a^2 instead of x^2

If that so, So here is the solution :

$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} }$

On rationalizing the denominator we get,

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


Squaring both the sides we get,

${(x + \frac{1}{x} )}^{2} = {(4)}^{2} \\ \\ {(x)}^{2} + {( \frac{1}{x} )}^{2} + 2 \times x \times \frac{1}{x} = 16 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 14$

Hope this helps!!!

Feel free to ask in the comment section if you have any doubt regarding to my answer...

@Mahak24

Thanks...
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