MHB How Do I Solve This Quadratic Inequality Correctly?

[tmt1]

I'm working on this problem.

$$2x^2 + 4x \ge x^2 - x - 6$$

I got here

$$2x -x \ge -3$$

But I don't know how to go from here.

 

[Jameson]

Hmm, care to show how you got there because I think you've made an error. The $x^2$ terms don't disappear in this problem. Try moving all the terms to the left hand side and tell me what you get. :)

 

[Lebec]

beginner, having a crack at problem

$$2x^2+4x\ge x^2-x-6$$
$$\implies x^2+5x+6\ge 0$$
$$\implies (x+2)(x+3)\ge 0$$
$$\implies x \ge-2 $$
$$ or \ x \ge -3$$

I have feeling that it should be:
$$ \ or \ x \le -3$$
but don't know the reason.
Could someone please explain?

 

[mathbalarka]

Your approach is correct until line #4. $(x + 2)(x + 3) \geq 0$ doesn't mean both of the factors $x + 2$ and $x + 3$ have to be $\geq 0$.

$a \cdot b \geq 0$ implies that either both of $a$ and $b$ are positive or both of $a$ and $b$ are negative (positive times negative is negative, so it can't be the case that one of $a, b$ is positive and another is negative).

Thus, either $x + 2 \geq 0, x + 3 \geq 0$ or $x + 2 \leq 0, x + 3 \leq 0$. In the first case we get, by rearranging, $x \geq -2, x \geq -3$ and in the second case, similarly, $x \leq -2, x \leq -3$.

Now $x \geq -2$ automatically implies $x \geq -3$, so knowing $x \leq -3$ is superfluous information : the relevant conclusion is $x \geq -2$.

Similarly, $x \leq -3$ automatically implies $x \leq -2$. So $x \leq -2$ is superfluous, and the information of interest is $x \leq -3$.

Hence, the inequality holds if and only if $x \geq -2$ or $x \leq -3$. In mathematical notation, the solution of the inequality is $x \in (-\infty, -3] \cup [-2, \infty)$.

 

[MarkFL]

$$2x^2+4x\ge x^2-x-6$$
$$\implies x^2+5x+6\ge 0$$
$$\implies (x+2)(x+3)\ge 0$$
$$\implies x \ge-2 $$
$$ or \ x \ge -3$$

I have feeling that it should be:
$$ \ or \ x \le -3$$
but don't know the reason.
Could someone please explain?

Yes, you are good at:

$$(x+2)(x+3)\ge0$$

We know the expression on the left is a parabola opening up, and having roots at $x=-3,-2$. Therefore it must be non-negative on:

$$(-\infty,-3]\,\cup\,[-2,\infty)$$