Question

If x^2-8x-1=0 then find the value of x^2+1/x^2

Answer 1

${x}^{2} - 1 = 8x \\ \frac{ {x}^{2} }{x} - \frac{1}{x} = 8 \\ x - \frac{1}{x } = 8 \\ squaring \: both \: sides \\ {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 64 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 66$

Answer 2

Given:

$x^{2} + \frac {1}{x^{2}}$

and $x^{2}-8x-1=0$

To find:

The value of $x^{2} + \frac {1} {x^{2}}$

Answer:

The given equations is $x^{2}-8x-1 = 0$

We need to find the $x^{2} + \frac {1} {x^{2}} = ?$

$x^{2}-8x-1 = 0$

$x^{2}-1 = 8x$

$\frac {x^{2}} {x} - \frac {1} {x} = 8$

$x - \frac {1}{x} = 8$

Squaring on both sides

${x- \frac {1}{x}}^{2}= 8^{2}$

$x^{2} + \frac {1}{x^{2}} - 2=64$

$x^{2} + \frac {1}{x^{2}}= 64 - 2$

$x^{2} + \frac {1}{x^{2}} = 62$