Question

If x=2+√3. find the value of x^2+1/x^2​

Answer 1

Answer:x = 2 + √3

1/x = 1/2 + √3

= 1 × (2 - √3)/(2 + √3) (2 - √3)

= (2 - √3)/(2^2 - √3^2)

= (2 - √3)/4 - 3

= (2 - √3)

Therefore ,

x^2 = (2 + √3)

= (2)^2 + (√3)^2 + 2 × 2 × √3

= 4 + 3 + 4√3

= 7 + 4√3

1/x^2 = (2 - √3)^2

= (2)^2 + (√3)^2 - 2 × 2 × √3

= 4 + 3 - 4√3

= 7 - 4√3

x^2 + 1/x^2

= (7 + 4√3) + (7 - 4√3)

= 7 + 4√3 + 7 - 4√3

= 7 + 7 + 4√3 - 4√3

= 14

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Answer 2

AnswEr :

• $\ \boxed{\boxed{\sf x^2 + \dfrac{1}{x^2} = 14}}$

Explanation :

Given that

$\sf \: x = 2 + \sqrt{3}$

Firstly,we need to calculate the value of 1/x.

Here,

$\sf \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3} }$

Rationalising the denominator

$\implies \: \sf \: \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \implies \: \sf \: \dfrac{1}{x} = \frac{2 - \sqrt{3} }{(2 + \sqrt{3})(2 - \sqrt{3} ) }$

Expanding the denominator : a² - b² = (a + b)(a - b)

Thus,

2² - (√3)²

= 4 - 3

= 1

Thus,

$\implies \: \sf \: \dfrac{1}{x} = \dfrac{2 - \sqrt{3} }{4 - 3} \\ \\ \implies \: \boxed{ \sf \: \frac{1}{x} = 2 - \sqrt{3} }$

Now,

$\sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } \\ \\ \longrightarrow \: \sf \: (2 + \sqrt{3} ) {}^{2} + (2 - \sqrt{3} ) {}^{2} \\ \\ \longrightarrow \: \sf \: (7 + 2 \sqrt{3} ) + (7 - 2 \sqrt{3} ) \\ \\ \longrightarrow \: 14$

The value of above expression is 14.