# If x2-8x-1=0 then prove that x2+1/x2 ?

Jun 21, 2018

${x}^{2} + \frac{1}{x} ^ 2 = \frac{64 {x}^{2} + 16 x + 2}{1 + 8 x}$

#### Explanation:

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

To Find: ${x}^{2} + \frac{1}{x} ^ 2$
Solution:
${x}^{2} - 8 x - 1 = 0$

$\implies {x}^{2} = 1 + 8 x$

$\implies \therefore \frac{1}{x} ^ 2 = \frac{1}{1 + 8 x}$

$\implies \therefore {x}^{2} + \frac{1}{x} ^ 2 = \left(1 + 8 x\right) + \frac{1}{1 + 8 x}$

$\implies {x}^{2} + \frac{1}{x} ^ 2 = \frac{\left(1 + 8 x\right) \left(1 + 8 x\right) + 1}{1 + 8 x}$

$= \frac{{\left(1 + 8 x\right)}^{2} + 1}{1 + 8 x} = \frac{1 + 16 x + 64 {x}^{2} + 1}{1 + 8 x}$

$= \frac{64 {x}^{2} + 16 x + 2}{1 + 8 x}$

So, ${x}^{2} + \frac{1}{x} ^ 2 = \frac{64 {x}^{2} + 16 x + 2}{1 + 8 x}$

Jun 21, 2018

The question looks incomplete, but I'll do whatever I can

#### Explanation:

Since we know that

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

we can isolate ${x}^{2}$ to get

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

So, ${x}^{2} + \frac{1}{x} ^ 2$ becomes

$\left(8 x + 1\right) + \frac{1}{8 x + 1} = \frac{{\left(8 x + 1\right)}^{2} + 1}{8 x + 1} = \setminus \frac{64 {x}^{2} + 16 x + 2}{8 x + 1}$

Now, I don't know what you want to prove, but for sure we have

${x}^{2} + \frac{1}{x} ^ 2 = \setminus \frac{64 {x}^{2} + 16 x + 2}{8 x + 1}$

Jun 22, 2018

${x}^{2} + \frac{1}{x} ^ 2 = 66$.

#### Explanation:

$\text{Given that, } {x}^{2} - 8 x - 1 = 0. \ldots \ldots \ldots . . \left(\ast\right)$.

$\text{Here, "x!=0, because, "if "x=0," then, }$

${x}^{2} - 8 x - 1 = 0 \Rightarrow 0 - 1 = 0 , \text{ which is not possible.}$

$\text{Hence, dividing "(ast)" throughout by "x!=0," we get, }$

$x - 8 - \frac{1}{x} = 0 , \mathmr{and} , x - \frac{1}{x} = 8$.

$\therefore {\left(x - \frac{1}{x}\right)}^{2} = {8}^{2} = 64$.

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

$\Rightarrow {x}^{2} + \frac{1}{x} ^ 2 = 64 + 2 = 66$.