Problem: Solve the equation

\(x^{2}+2 a x+\displaystyle\frac{1}{16}=-a+\sqrt{a^2+x-\displaystyle\frac{1}{16}}\)for \(x\) real and \(0alt : plots

There is an easy way to attack the solution of this equation and a hard way. The hard way is to add [latex "size="-2"]a\) to both sides and square, generating a fourth degree equation in \(x\). No thanks. The easy way, however, is to note that the right-hand side is a solution to the equation

\(x=y^{2}+2 a y+\displaystyle\frac{1}{16}\)Furthermore, the solution to this equation gives

\(y=x^{2}+2 a x+\displaystyle\frac{1}{16}\).

That is, we have an equation for points where a function equal to its inverse function. The points in such a case lie along the line \(y=x\) (as can be verified in the plot), and we simply have a quadratic to solve:

\(x=x^{2}+2 a x+\displaystyle\frac{1}{16}\),

the solution to which is

\(x=\displaystyle\frac{1}{2}-a \pm \sqrt{\displaystyle\frac{3}{4}-4 a (1-a)}\).