Question

If x = 2 +√3 then find the value of x² + 1/x²

Answer 1

ANSWER:

• Value of x²+1/x² = 14

GIVEN:

• x = 2+√3

TO FIND:

• Value of x² + 1/x².

SOLUTIOÑ:

=> x = 2+√3

$\implies \: \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3} } \\ \\ \implies \: \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \implies \: \dfrac{1}{x} = \dfrac{2 - \sqrt{3} }{4 - 3} \\ \\ \implies \: \dfrac{1}{x} = 2 - \sqrt{3}$

Now:

$= x {}^{2} + \dfrac{1}{x {}^{2} } \\ \\ = (2 + \sqrt{3} ) {}^{2} + (2 - \sqrt{3} ) {}^{2} \\ = 4 + 3 + 4 \sqrt{3} + 4 + 3 - 4 \sqrt{3} \\ = 14$

Value of x²+1/x² = 14

Answer 2

$\huge\mathfrak{Answer:}$

Given:

• We have been given that x = 2 +√3 .

To Find:

• We need to find the value of x² + 1/x².

Solution:

It is given that x = 2 + √3

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

$\implies\sf{ \dfrac{1}{x} = \dfrac{2 - \sqrt{3} }{ {2}^{2} - { (\sqrt{3} )}^{2} } }$

[Using (a - b)(a + b) = a² - b²]

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

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

Now,

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

$\implies\sf{x + \dfrac{1}{x} = 4}$

Now, we need to find the value of x² + 1/x², we have

$\sf{ {(x + \dfrac{1}{x}) }^{2} = ( {4}^{2} )}$

$\implies\sf{ {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 16}$

$\implies\sf{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 16 - 2}$

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

Hence, the value of x² + 1/x² is 14.