\(x + \frac{1}{x} = 2 \\ x + \frac{1}{x} - 2 = 0 \\ \frac{x^2}{x} + \frac{1}{x} - \frac{2x}{x} = 0 \\ \frac{x^2+1-2x}{x} = 0\)
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
\(1^2 + \frac{1}{1^2} = 1 + 1 = 2\)
.
