Question

x=3+root 8 find value of x^2 +1/x^2​

Answer 1

Required Answer:-

Given:

• $\sf x = 3 + \sqrt{8}$

To Find:

• $\sf {x}^{2} + \dfrac{1}{ {x}^{2} } = \: ?$

Solution:

We have,

$\sf \implies x = 3 + \sqrt{8}$

Therefore,

$\sf \implies \dfrac{1}{x} = \dfrac{1}{3 + \sqrt{8} }$

$\sf \implies \dfrac{1}{x} = \dfrac{1}{3 + \sqrt{8} } \times \dfrac{3 - \sqrt{8} }{3 - \sqrt{8} }$

$\sf \implies \dfrac{1}{x} = \dfrac{3 - \sqrt{8} }{ {(3)}^{2} - {( \sqrt{8} )}^{2} }$

$\sf \implies \dfrac{1}{x} = \dfrac{3 - \sqrt{8} }{ 9 - 8 }$

$\sf \implies \dfrac{1}{x} = 3 - \sqrt{8}$

Therefore,

$\sf \implies x + \dfrac{1}{x} = 3 + \sqrt{8} + 3 - \sqrt{8}$

$\sf \implies x + \dfrac{1}{x} = 6$

Squaring both sides, we get,

$\sf \implies \bigg(x + \dfrac{1}{x} \bigg)^{2} = 36$

$\sf \implies {x}^{2} + \dfrac{1}{ {x}^{2}} + 2 \times x \times \dfrac{1}{x} = 36$

$\sf \implies {x}^{2} + \dfrac{1}{ {x}^{2}} + 2 = 36$

$\sf \implies {x}^{2} + \dfrac{1}{ {x}^{2}} = 34$

★ Hence, the value of x² + 1/x² is 34.

Answer:

• $\sf{x}^{2} + \dfrac{1}{ {x}^{2}} = 34$

Identity Used:

• (a + b)² = a² + 2ab + b²