$\qquad \qquad \qquad$ 
I would like to know if my opinion about this solution is correct.
The solutions $x = 0$ and $x = -\frac 43$ are correct if and only if (iff) such that $b=- \frac x2 \Longrightarrow \sqrt{x^2+x}=- \frac x2$ holds. Because, the variables $"a"$ and $"b"$ are dependent. They are not independent variables. So, the solution must be if and only if
$$\begin{cases} \displaystyle \sqrt{x^2+1}=\dfrac{2-x}{2} \\ \sqrt{x^2+x}=- \dfrac x2 \end{cases}$$
Is my claim correct?
And here are my attempts:
$$\sqrt{x^2+1}+\sqrt{x^2+x}=1-x \\ \left(\sqrt{x^2+1}+\sqrt{x^2+x}\right) \left(\sqrt{x^2+1}-\sqrt{x^2+x}\right)=(1-x)\left(\sqrt{x^2+1}-\sqrt{x^2+x}\right) \\ 1-x=(1-x)\left(\sqrt{x^2+1}-\sqrt{x^2+x}\right) \\ \sqrt{x^2+1}+\sqrt{x^2+x}=1, x\neq 1 \\ 2\sqrt{x^2+x}=-x \Longrightarrow x_1=0, x_2=- \frac 43 \\ 2\sqrt{x^2+1}=2-x \Longrightarrow x_1=0, x_2=-\frac 43 $$