How do I solve this equation for x? $x=(1-\frac{1}{x})^\frac{1}{2}+(x-\frac{1}{x})^\frac{1}{2}$
How do I solve for x?

We see that
\begin{align}
x&=(1-\tfrac1x)^\frac12+(x-\tfrac1x)^\frac12\\
x^2&=1-\frac1x+x-\frac1x+2(1-\tfrac1x)^\frac12(x-\tfrac1x)^\frac12\\
x^2-x+\frac2x-1&=2(1-\tfrac1x)^\frac12(x-\tfrac1x)^\frac12\\
\left(x^2-x+\frac2x-1\right)^2&=4(1-\tfrac1x)(x-\tfrac1x)\\
-3+\frac4{x^2}-\frac4x+6x-x^2-2x^3+x^4&=-4+\frac4{x^2}-\frac4x+4x\\
0&=x^4-2x^3-x^2+2x+1\\
0&=(1+x-x^2)^2\\
0&=x^2-x-1\\
\end{align}
And so we see that $x=\phi$ or $x=1-\phi$, the golden ratio. We need to check our solutions however, since squaring can create extra ones; and we see, that only $x=\phi$ satisfies the original equation.
