$ 1 + \sqrt{1 - \sqrt{x^4-x^2}} = x$ can be written in the form $\frac{a}{b}$, find $a+b$. The solution to the equation 
$$1 + \sqrt{1 - \sqrt{x^4-x^2}} = x$$
 can be written in the form $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers. Find $a+b$.

First isolate the square root on one side:
$$\sqrt{1-\sqrt{x^4-x^2}}=x-1\;.$$
Now square:
$$1-\sqrt{x^4-x^2}=(x-1)^2=x^2-2x+1\;.$$
Then $-\sqrt{x^4-x^2}=x^2-2x$, $\sqrt{x^4-x^2}=2x-x^2$, and $x^4-x^2=(2x-x^2)^2=x^4-4x^3+4x^2$. Simplification yields the equation $4x^3-5x^2=0$, or $x^2(4x-5)=0$. This is easy to solve. Be careful, though: you need to check each solution back in the original equation, because the squaring steps may have introduced extraneous solutions.
