# Thread: Quadratic Compound Inequality Question

1. ## Quadratic Compound Inequality Question

Good morning all!

I am studying to prepare myself for a PreCalc/Trig class in the Fall and I am stuck on a quadratic compound inequality with no examples to use and no professor to ask. I have attached a copy of the work I have done on this exercise so far. Would any of you guys mind letting me know where I went wrong ?

Questions I had while solving:

When I derived the two possible values for the absolute value, should I have eliminated part of the domain since it has both positive and negative 3? If so, why?

Is there a better way I can analyze the domains I found to make it easier to find the intersections?

Any help would be really appreciated.

Lenora

2. ## Re: Quadratic Compound Inequality Question

Beer soaked suggestion follows.
Originally Posted by lenoradgray
Good morning all!

I am studying to prepare myself for a PreCalc/Trig class in the Fall and I am stuck on a quadratic compound inequality with no examples to use and no professor to ask. I have attached a copy of the work I have done on this exercise so far. Would any of you guys mind letting me know where I went wrong ?

Questions I had while solving:

When I derived the two possible values for the absolute value, should I have eliminated part of the domain since it has both positive and negative 3? If so, why?

Is there a better way I can analyze the domains I found to make it easier to find the intersections?

Any help would be really appreciated.

Lenora
One easy way is to plot all 3 functions on the same window; you can easily see the solution set to this quadratic compound inequality at
plot |x^2-9|, 2, 9, x=-5 to 5 - Wolfram|Alpha Results
Too hammered and rusty at the moment but I'm sure you can verify this easily. Check your results at
solve 2<=|x^2-9|<9 - Wolfram|Alpha Results

3. ## Re: Quadratic Compound Inequality Question

well, you picked a rather tedious one to solve ...

$2 \le |x^2-9| < 9$

note ... $|u| = \sqrt{u^2}$

part (1) ...

$|x^2-9| \ge 2$

$\sqrt{(x^2-9)^2} \ge 2$

$(x^2-9)^2 \ge 4$

$(x^2-9)^2 - 4 \ge 0$

$[(x^2-9)-2][(x^2-9)+2] \ge 0$

$(x^2-11)(x^2-7) \ge 0$

for the inequality to be true, both factors must be non-negative or both must be strictly negative.

both factors are non-negative for $x \ge \sqrt{11}$ or $x \le -\sqrt{11}$

both factors are negative for $-\sqrt{7} < x < \sqrt{7}$

solution for this part of the inequality is the union of the above intervals.

part (2) ...

$|x^2-9| < 9$

$\sqrt{(x^2-9)^2} < 9$

$(x^2-9)^2 < 81$

$(x^2-9)^2 - 81 < 0$

$[(x^2-9)-9][(x^2-9)+9] < 0$

$(x^2-18)(x^2) < 0$

since $x^2 \ge 0$ for all $x$, the above inequality is true for $(-\sqrt{18} < x < 0) \cup (0 < x < \sqrt{18})$

so, the solution set of the inequality $2 \le |x^2-9| < 9$ is the intersection of the intervals

$(x \ge \sqrt{11}) \cup (x \le -\sqrt{11})$

$-\sqrt{7} < x < \sqrt{7}$

$(-\sqrt{18} < x < 0) \cup (0 < x < \sqrt{18})$

which is ...

$(-\sqrt{18} < x \le -\sqrt{11}) \cup (-\sqrt{7} \le x < 0) \cup (0 < x \le \sqrt{7}) \cup (\sqrt{11} \le x < \sqrt{18})$

4. ## Re: Quadratic Compound Inequality Question

Thanks so much for your help!