XII Moscow State University Mathematical Olympiad, Problem 8

Find all real solutions of the equation

x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}},   0 < a < \frac{1}{4}.

Solution

Introduce y = x^2+2ax+\frac{1}{16}. Then squaring the right-hand side gives x = y^2+2ay+\frac{1}{16}. But

y = x^2+2ax+\frac{1}{16}

and

x = y^2+2ay+\frac{1}{16}

are inverse function so that their graphs are reflections of each other in the main diagonal y = x. Solutions to the original equation lie on both graphs and, hence, on the diagonal. It follows that any such solution also satisfies

x = x^2+2ax+\frac{1}{16}.

This is a quadratic equation

x^2+(2a-1)x+\frac{1}{16}=0,

with two solutions

x_{1,2}=\frac{-(2a-1) \pm \sqrt{(2a-1)^2-\frac{1}{4}}}{2}.

For 0 < a < \frac{1}{4}, -1 < (2a-1) < -\frac{1}{2}, so that

(2a-1)^2 > \frac{1}{4}

which means that the expression under the radical is positive and the roots are real.

To make the verification easier set b = \frac{1-2a}{2} and observe that then a = \frac{1-2b}{2}. In this notations,

x_{1,2}=b \pm \sqrt{b^2-\frac{1}{16}}.