How many real solutions of this equation?

Question

How many real solutions of this equation $$\sqrt{x}+\sqrt{1-x^2}=\sqrt{2-3 x-4 x^2}?$$ I posted my question at here https://mathematica.stackexchange.com/questions/51316/how-can-i-get-the-exact-real-solution-of-this-equation In my opinion, there is one solution is $\frac{1}{9}\left(\sqrt{34}-5\right).$

Answer 1

$$\sqrt{1 - x^2} + \sqrt{x} = \sqrt{-4x^2 - 3x + 2}$$ $$2\sqrt{x}\sqrt{1 - x^2} - x^2 + x + 1 = -4x^2 - 3x + 2$$ $$2\sqrt{x}\sqrt{1 - x^2} = -3x^2 - 4x + 1$$ $$4x - 4x^3 = 9x^4 + 24x^3 + 10x^2 - 8x + 1$$

Has solutions: $$x \in \left\{-\frac{\sqrt{34} + 5}{9}, \frac{\sqrt{34} - 5}{9}, -\sqrt{2} - 1, \sqrt{2} - 1\right\}$$

Checking the four values, 2 are valid:

$$x \in \left\{\frac{\sqrt{34} - 5}{9}, -\sqrt{2} - 1\right\}$$

Note that the second value extends the range of the square root into complex numbers.

Answer 2

If you expand and eliminate the square roots in the obvious way, you get a quartic which factorises as $$(9x^2+10x-1)(x^2+2x-1)\ .$$ This has four real roots, but they may not all work in your original equation - you will need to check.