System of inequalities $x^2+2x+\alpha\leq0$ and $x^2-4x-6\alpha\leq 0$ has unique solution. 
Find all values of $\alpha$ for which the system of inequalities $x^2+2x+\alpha\leq0$
and $x^2-4x-6\alpha\leq 0$
has unique solution.

My Try: We can write it as $$\frac{x^2-4x}{6}\leq \alpha \leq -x^2-2x$$
So we get $$\frac{(x-2)^2-4}{6}\leq \alpha \leq -(x+1)^2+1$$
So for the existence of solution, Here $-\frac{2}{3}\leq \alpha \leq 1$.
Now how can I solve for unique solution?

As both the Quadratic inequalities is less than zero, so we can say that their discriminant must be greater than or equal to Zero ,i.e., real roots must exists.
So, we will get $$4-4\alpha \ge 0$$ $$16 + 24\alpha \ge 0$$On solving these eqations, we will get $$\frac{-2}3 \le \alpha \le 1$$ Now, we can determine the roots of these quadratic equations as, $$-1 \pm \sqrt{1-\alpha}$$ $$2 \pm \sqrt{4 + 6\alpha}$$ Now, we can see that both the roots of first equation are negative and one of the roots of the other equation is positive.
Therefore for a unique solution, the other root of the second equation must be negative. Thus, $$2 - \sqrt{4 + 6\alpha} \lt 0$$ On solving this, we get$$\alpha \gt 0$$ and we know that, $$\frac{-2}3 \le \alpha \le 1$$ Therefore, the required set of values of $\alpha$ is $$0 \lt \alpha \le 1$$
