Mathematical Reflections 2017, Issue 3 - Problem S409

Problem:
Solve in real numbers the equation $$2\sqrt{x-x^2}-\sqrt{1-x^2}+2\sqrt{x+x^2}=2x+1.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Clearly, $0 \leq x \leq 1$. Observe that
$$\renewcommand{\arraystretch}{2} \begin{array}{rcl} (\sqrt{x}+\sqrt{1-x})^2&=&1+2\sqrt{x-x^2} \\ \dfrac{1}{2}(\sqrt{1-x}+\sqrt{1+x})^2&=&1+\sqrt{1-x^2} \\ (\sqrt{x}-\sqrt{1+x})^2&=&2x+1-2\sqrt{x+x^2}. \end{array}$$
So, the given equation is equivalent to
$$(\sqrt{x}+\sqrt{1-x})^2-(\sqrt{x}-\sqrt{1+x})^2=\dfrac{1}{2}(\sqrt{1-x}+\sqrt{1+x})^2,$$
i.e. $$(\sqrt{1-x}+\sqrt{1+x})(2\sqrt{x}+\sqrt{1-x}-\sqrt{1+x})=\dfrac{1}{2}(\sqrt{1-x}+\sqrt{1+x})^2.$$
Since $\sqrt{1-x}+\sqrt{1+x}>0$, then the equation becomes $$2\sqrt{x}+\sqrt{1-x}-\sqrt{1+x}=\dfrac{1}{2}(\sqrt{1-x}+\sqrt{1+x}),$$ i.e. $$4\sqrt{x}=3\sqrt{1+x}-\sqrt{1-x}.$$ Squaring both sides, we get $$16x=9(1+x)+(1-x)-6\sqrt{1-x^2},$$ i.e. $$(1+x)+9(1-x)-6\sqrt{1-x^2}=0,$$ so $$\left(\sqrt{1+x}-3\sqrt{1-x}\right)^2=0.$$ It follows that $\sqrt{1+x}=3\sqrt{1-x}$, i.e. $1+x=9(1-x)$, which gives $x=\dfrac{4}{5}$.