TITLE: Solve $x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $
QUESTION [8 upvotes]: Find the real roots of the equation $$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $$ $$(0<a<\frac{1}{4})$$

My attempt,
$(x^2+2ax+\frac{1}{16}+a)^2=a^2+x-\frac{1}{16}$
I did an expansion which becomes  $x^4+4ax^3(4a^2+2a+\frac{1}{8})x^2+(4a^2+\frac{1}{4}a-1)x+\frac{1}{8}a+\frac{1}{16}+\frac{1}{256}=0$
$$\left [ x^4+(2a-1)x^3+\frac{1}{16}x^2 \right ]+\left [ (2a-1)x^3+(4a^2-1)x^2+(\frac{a}{8}+\frac{1}{16})x \right ] +\left [ (2a+\frac{17}{16})x^2+(4a^2+\frac{1}{8}a-\frac{17}{16})x+(\frac{1}{8}a+\frac{1}{16}+\frac{1}{256}) \right ]=0$$
$$\left [ x^2+(2a-1)x+\frac{1}{16} \right ]\left [ x^2(2a+1)x+(2a+\frac{17}{16}) \right ]=0$$
Since $0<a<\frac{1}{4}$, 
$x^2+(2a-1)x+\frac{1}{16}=0$
$x=\frac{1-2a}{2}\pm \sqrt{(\frac{1-2a}{2})^2-\frac{1}{16}}$
My solution is really messy and I'm unsure my answer.
My question: is my answer correct? And is there another way to solve this algebra question? Thanks in advance.

REPLY [5 votes]: Hint
$$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2-a^2+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2=a^2-\frac{1}{16}-a+\sqrt{a^2+x-\frac{1}{16}}\\
(x+a)^2+(x+a)=a^2+x-\frac{1}{16}+\sqrt{a^2+x-\frac{1}{16}}$$
Now call:
$$y=\sqrt{a^2+x-\frac{1}{16}}\to y^2=a^2+x-\frac{1}{16}$$
and then
$$(x+a)^2+(x+a)=y^2+y\\
(x+a)^2-y^2+(x+a)-y=0\\
(x+a-y)(x+a+y+1)=0$$
Can you finish?