# Homework Help: Absolute value with quadratic

1. Jun 25, 2011

### flyingpig

1. The problem statement, all variables and given/known data

Very embarrassing that I had problems with this. I managed to solved it at the end, but I am posting this for two reasons

1) I like someone else to work it out in a different method, I wnt to see what other approaches there are

2) Whether mine is taking too long or not. I mean I solved it, but any unnecessary steps?

3) Check whether I am right or wrong lol

$$|x^2 + 6x + 16| < 8$$

3. The attempt at a solution

At first I was going to do

$$- 8 < x^2 + 6x + 16 < 8$$

Then I realize it was hopeless

So then I did

$$(x^2 + 6x + 16) < 8$$ and $$- ( x^2 + 6x + 16 ) < 8$$

(1) $$(x^2 + 6x + 16) < 8$$

$$x^2 + 6x + 8 < 0$$

$$(x+4)(x+2) < 0$$

Did some test points and found that $$x \in (-4,-2)$$ is a solution

(2) $$-(x^2 + 6x + 16) < 8$$

$$-(x+4)(x+2) < 0$$

$$(x+4)(x+2) > 0$$

Now here is the problem should i have even divide that -1 and switch the inequality signs? I could and I would get some meaningless answer like $$x \in (-\infty,-4)$$

Anyways I threw it back in

$$(-x-4)(x+2) > 0$$

x is still 4, so no change, neither did the solution

So my solution remains as $$x \in (-4,-2)$$

2. Jun 25, 2011

### ehild

The second line is wrong.

ehild

3. Jun 25, 2011

### Mentallic

This isn't hopeless. Solve both cases separately, mainly:

$$-8<x^2+6x+16$$
$$x^2+6x+16<8$$
and then find the intersection of each solution set.

This is equivalent to

And you can improve on this part

By simply drawing a rough sketch of the quadratic (you should be able to visualize it in your head even).

And ehild has pointed out your error.
What you did is basically say that since a<b (from your first inequality) then a-b<0, but if -a<b, then -a-b<0 but -a-b is not equal to -(a-b) as you did.

One more point: Even though you solved the wrong quadratic,

This would not be true since you need to find the intersection of the solution sets $x\in(-\infty,-4)$ and $x\in(-4,-2)$ which is the empty set (x cannot be in both at the same time, so there is no x satisfying these solution sets).

4. Jun 25, 2011

### flyingpig

But I thought

|x| < a

is

-x<a and x <a

A definition of piecewise?

5. Jun 25, 2011

### Mentallic

Look more closely, that's not what the problem was.

6. Jun 25, 2011

### Staff: Mentor

Or equivalently, x > -a and x < a. This is normally written as -a < x < a, assuming a is a positive number.
???
I don't understand your thinking here.

7. Jun 25, 2011

### SammyS

Staff Emeritus
To take what ehild wrote, one or two steps farther;

If $- ( x^2 + 6x + 16 ) < 8\,,$ then $-(x^2+6x+24)<0\,.$

Now solve that.

8. Jun 25, 2011

### SammyS

Staff Emeritus
You might try completing the square for the quadratic, x2 + 6x + 16 .

x2 + 6x + 16 = x2 + 6x + 9 + 7
= (x + 3)2 + 7​

For on thing, this tells you that the minimum value for the quadratic is 7.

9. Jun 26, 2011

### flyingpig

But there aer no real solutions for x^2 + 6x + 24

10. Jun 26, 2011

### ehild

That means, it never crosses the x axis. It is either positive or negative everywhere. Which one?

ehild

11. Jun 26, 2011

### Mentallic

Notice the post just before yours:

12. Jun 26, 2011

### flyingpig

Yeah so you can't solve it, it's always positive and never less than 0.

13. Jun 26, 2011

### Mentallic

If we're trying to solve $-(x^2+6x+24)<0$ this is equivalent to solving $x^2+6x+24>0$

14. Jun 26, 2011

### ehild

So −8<x2+6x+16 is always true.

ehild

15. Jun 26, 2011

### flyingpig

Oh okay so that means all values of x will work, but doesn't this contradict my answer that (-4,-2)

16. Jun 26, 2011

### ehild

No, the other side of the inequality holds only in the interval (-4,-2)

ehild

17. Jun 26, 2011

### Mentallic

x has to be satisfied by both solution sets at the same time. It's called taking the intersection.

18. Jun 26, 2011

### SammyS

Staff Emeritus
Well, $x^2+6x+16$ is positive for all x.

What does that tell you about $\left|x^2+6x+16\right|\,?$

19. Jun 26, 2011

### flyingpig

Not changing...

20. Jun 26, 2011

### rock.freak667

So if |x2+6x+16| means to take the positive value of "x2+6x+16" and "x2+6x+16" is always positive, then you do you really need the modulus sign?