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?
 A: 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 $$0 \lt \alpha \le 1$$
A: These are two parabolas pointing up. Both are $\leq 0$ in some interval $[x_1, x_2]$ where these are the real roots of the respective parabola (if real roots exists of course, we want them to exist, even if they coincide, hence the restrictions you found for $\alpha$ guarantee this, OK).  
Now graph some rough drawing. For this system to have a unique solution, the bigger root of one of the the parabolas has to be equal to the smaller root of the other parabola (if that's no so, the intersection of these two closed intervals will be either empty /no solutions to the system/, or a non-degenerate interval /infinitely many solutions/).    
So you get:    
$x^2+2x+\alpha = (x-b)(x-c)$
$x^2-4x-6\alpha = (x-b)(x-d)$     
And also:  $d \leq b \leq c$ or $c \leq b \leq d$.    
A: A way that maybe can help you to understand completely what you want to know, is enter to Desmos Calculator and mark "slider control" for the number $\alpha$ ($a$ in the figure, for the value $a=0$). This allow you to visualize what happen with variation of $\alpha$.

