Common solutions of two inequations

Question

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.

What is the basis of assumption of the existence of a real root ? . Anyways, both equations are upward facing parabolas. Can you re-structure the equations so that the vertex of the parabolas become apparent ? — Shailesh, Jul 27, 2015 at 5:11
Note that you cannot subtract inequalities in general. Subtraction is equivalent to adding the negative and adding. As you may know, taking negatives reverses the inequality, hence such a step is not valid. — Macavity, Jul 27, 2015 at 5:11

Macavity · Accepted Answer · 2015-07-27 05:52:41Z

2

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 $a \in \{0, 1\}$.

answered Jul 27, 2015 at 5:52

Macavity

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$\begingroup$ Are the values of $a\epsilon {{0,1}}$,guess work or there is some method to find them out from equations (A) and (B) $\endgroup$
– Vinod Kumar Punia
Jul 27, 2015 at 6:03
$\begingroup$ I thought you wanted hints. The only ways there is a singleton intersection for two intervals is when either there is overlap only at a boundary, or if one of the intervals is itself a singleton and included in the other... $\endgroup$
– Macavity
Jul 27, 2015 at 6:07

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