<s>[INST] Common solutions of two inequations Find the real values of $a$ for which the inequations $x^2-4x-6a\leq 0$ and $x^2+2x+a\leq0$ have only one real solution common.
My attempt:
Let $\alpha$ be one real common root of two inequations.
then $ \alpha^2-4\alpha-6a\leq 0$.......(1)
$\alpha^2+2\alpha+a\leq 0$.......(2)
Subtract the two inequations,we get
$-6\alpha-7a\leq0$ but could not solve further for values of $a$.
$\alpha^2-4\alpha-6a\leq 0$ can be written as $(\alpha-2)^2-6a-4\leq0$
$\alpha^2+2\alpha+a\leq 0$ can be written as $(\alpha+1)^2+a-1\leq0$
Any hint will be helpful for me.
 [/INST] Hint: 
$$x^2-4x-6a \le 0 \iff 2-\sqrt{4+6a} \le x \le 2+\sqrt{4+6a} \tag{A}$$
$$x^2+2x+a \le 0 \iff -1-\sqrt{1-a} \le x \le -1+\sqrt{1-a} \tag{B}$$
Now find the situations when $A \cap B$ is exactly one point, to get $a \in \{0, 1\}$.
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