# Tough equation, easy solution

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

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

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