An useful question

if x+ 1/x= 2, what is x2 + 1/x2

if  $x+ \frac{1}{x}= 2$ , what is $x^2 + \frac{1}{x^2}$

$\begin{array}{|rcll|} \hline x+ \frac{1}{x} &=& 2 \quad & | \quad \text{square both sides} \\ \left(~x+ \frac{1}{x}~\right)^2 &=& 2^2 \\ \left(~x+ \frac{1}{x}~\right)^2 &=& 4 \\ x^2 +2\cdot x\cdot \frac{1}{x} + \frac{1}{x^2} &=& 4 \\ x^2 +2 + \frac{1}{x^2} &=& 4 \quad & | \quad -2 \\ x^2 + \frac{1}{x^2} &=& 4-2 \\ \mathbf{x^2 + \frac{1}{x^2}} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}$

$x + \frac{1}{x} = 2 \\ x + \frac{1}{x} - 2 = 0 \\ \frac{x^2}{x} + \frac{1}{x} - \frac{2x}{x} = 0 \\ \frac{x^2+1-2x}{x} = 0$

We want to know what makes the numerator = 0.

You can also just say multiply both sides by x.

$x^2+1-2x = 0 \\ (x-1)(x-1) = 0 \\ x = 1$

(You can easily test this and see that 1 + 1/1 = 2)

So

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