Absolute Value with Quadratic: What Are the Solutions to |x^2 + 6x + 16| < 8?

[flyingpig]

Homework Statement

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

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

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 $$x \in (-4,-2)$$

 

[ehild]

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

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

The second line is wrong.


ehild

 

[Mentallic]

At first I was going to do

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

Then I realize it was hopeless

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

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

And you can improve on this part

(x+4)(x+2)<0Did some test points and found that
x∈(−4,−2)

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,

So my solution remains as
x∈(−4,−2)

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).

 

[flyingpig]

The second line is wrong.


ehild

But I thought

|x| < a

is

-x<a and x <a

A definition of piecewise?

 

[Mentallic]

But I thought

|x| < a

is

-x<a and x <a

A definition of piecewise?

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

 

[Mark44]

But I thought

|x| < a

is

-x<a and x <a

Or equivalently, x > -a and x < a. This is normally written as -a < x < a, assuming a is a positive number.

A definition of piecewise?

?
I don't understand your thinking here.

 

[SammyS]

The second line is wrong.

ehild

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.

 

[SammyS]

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.

 

[flyingpig]

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

 

[ehild]

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

ehild

 

[Mentallic]

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

Notice the post just before yours:

= (x + 3)² + 7

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

 

[flyingpig]

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

 

[Mentallic]

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

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

 

[ehild]

So −8<x²+6x+16 is always true.

ehild

 

[flyingpig]

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

 

[ehild]

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

ehild

 

[Mentallic]

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

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

 

[SammyS]

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

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

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

 

[flyingpig]

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

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

Not changing...

 

[rock.freak667]

Not changing...

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?

 

[flyingpig]

No...