Segment measure in right triangles

Question

The $\triangle ABO$ is rectangle in $O$, $OA=a$ and $OB=b$, if $AP=PQ=QB=x$, then $x=?$

I tried pythagoras in $\triangle POQ$, because $OP=a-x$, $OQ=b-x$ and $PQ=x$ and got a messy equation that i can't solve. That's what i tried for now.

Answer 1

Of course you can solve it easily, this being a quadratic. Some terms even cancel out nicely: $$(a-x)^2+(b-x)^2=x^2$$ $$x^2-2(a+b)x+a^2+b^2=0$$ $$x=\frac{2(a+b)\pm\sqrt{4(a+b)^2-4(a^2+b^2)}}2$$ $$x=a+b-\sqrt{2ab}$$ We reject $+$ in the last line because $PQ$ can never be longer than $a+b$.