# Absolute value with Quadratic Inequality

This is the problem:

$|x^2+2x-9|≤6$

I have no idea how to even begin with this, I'm really interested in how I should go about solving this inequality. Any help would be appreciated.

• If you can't solve it, try to draw a plot, when it's easy to see the all picture. – openspace Apr 16 '16 at 11:58

Variant: Rewrite $-6\le x^2+2x-9\le 6$ as \begin{align*}-6\le (x+1)^2-10\le 6&\iff 4\le (x+1)^2 \le 16\iff 2\le \lvert x+1\rvert\le4\\ &\iff x+1\in[-4,-2]\cup[2,4]\iff x\in[-5,-3]\cup[1,3]\end{align*}

• This is very nice indeed. +1 – DonAntonio Apr 16 '16 at 13:44

Fill in details and complete the proof:

$$|x^2+2x-9|\le6\iff -6\le x^2+2x-9\le 6$$

and now solve both inequalities and take the intersection* of both solution sets. For example

$$-6\le x^2+2x-9\iff x^2+2x-3\ge0\iff (x+3)(x-1)\ge0$$

and you get $\;x\le -3\;$ , or $\;x\ge 1\;$ . Take it from here.

• How did you get $x ≤ -3$ and $x≤ 1$ shouldn't the inequality signs be the other way round? – Daniel Brown Apr 16 '16 at 12:16
• @DanielBrown Not the first one, the second one is a typo. Thanks, editing. – DonAntonio Apr 16 '16 at 12:31