Question

if x-1/6=√6 than x2+1/x2 (A)2 (B)4 (C)6 (D)8​

Answer 1

(D)

Correction in question:

$x + \dfrac{1}{x} = \sqrt{6}$

$\sf Find \: \: {x}^{2} + \dfrac{1}{ {x}^{2} }$

Step-by-step explanation:

How to Solve ?

This is a very easy question, just follow two step

1) Square both side,

2) Then, Solve it simple

Easy ?

Solution:

$x - \frac{1}{x} = \sqrt{6}$

$(x - \dfrac{1}{x} ) {}^{2} = ( \sqrt{6} {)}^{2} \\ \\ {x}^{2} - 2.x. \frac{1}{x} + \frac{1}{ {x}^{2} } = 6 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 6 + 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 8$

Our answer is 8

Answer 2

Answer:

x-1/6=$\sqrt{6}$ then x =1/6+$\sqrt{6}$

x^2+1/x^2 therefor x^2 = (1/6+$\sqrt{6}$ )^2 =18       (36=$\sqrt{6}$ )

therfore 18+1/18 =12....

Step-by-step explanation: