Solving Quadratic Inequalities

[mathdad]

Section 2.6
Question 36Solve the quadratic inequality.

x^4 - 25x^2 + 144 ≤ 0

Can someone get me started?

 

[MarkFL]

Section 2.6
Question 36Solve the quadratic inequality.

x^4 - 25x^2 + 144 ≤ 0

Can someone get me started?

Factor:

$$(x^2-9)(x^2-16)\le0$$

Continue factoring...:D

 

[mathdad]

Thank you very much. I am at the AMC about to watch the shark movie 47 Meters Down. I will answer this and the other 3 questions later this evening. Thank you again for your continual help in this website.

 

[MarkFL]

Thank you very much. I am at the AMC about to watch the shark movie 47 Meters Down. I will answer this and the other 3 questions later this evening. Thank you again for your continual help in this website.

I stopped going to movie theaters several years ago because all the phones lit up like flashlights in my eyes during the entire show is just too distracting for me. :D

 

[mathdad]

I dislike movie theatres in NYC. I rarely go to the movies because of extremely overcrowded conditions and rude people constantly moving pass me on their way to the bathroom. I really want to see this movie because I find sharks to be interesting creatures.

- - - Updated - - -

 

[mathdad]

(x^2 - 9)(x^2 - 16) ≤ 0

(x - 3)(x + 3)(x - 4)(x + 4) ≤ 0

I can see right away that we must reject the follow values of x: 3, -3, 4, -4.

Our number line:

<---------(-4)------(-3)-----(3)-----(4)-------->

I will test each interval for algebra practice.

For (-infinity, -4), let x = -5. False statement for sure.

For (-4, -3), let x = -2. False statement.

For (-3, 3), let x = 0. False statement.

For (3, 4), let x = 3.5. True statement.

For (4, infinity), let x = 6. False statement.

The only solution is found in the interval (3, 4).

Correct?

Note: Most of the algebra is done on paper.

 

[MarkFL]

The solution is:

$$[-4,-3]\,\cup\,[3,4]$$

You missed an interval because -2 is not in the interval [-4,-3]... (an alarm should have gone off in your head when you saw the signs did not alternate while all roots are of odd multiplicity) and we include the end-points because the inequality is weak, and they do not cause division by zero.

The given expression is an even function, and so we should expect all behavior to be symmetrical across the $y$-axis. ;)

 

[mathdad]

I rushed through this problem. I can say silly errors were made but had this been a bonus question on a test, it would be a serious error. I will try to be more careful when solving each problem using the keyboard. If I had to work it out again, the silly errors would surely not be made.

 

[mathdad]

(x^2 - 9)(x^2 - 16) ≤ 0

(x - 3)(x + 3)(x - 4)(x + 4) ≤ 0

Setting each factor to 0, we get x = 3, -3, 4, and -4.

Our number line:

<---------(-4)------(-3)-----(3)-----(4)-------->

When x = -4, we get a true statement. The same can be said for x = -3, 3, and 4. This means they are part of the solution.

I will test each interval AGAIN for algebra practice.

For (-infinity, -4), let x = -5. False statement for sure.

For (-4, -3), let x = -3.5. True statement.

For (-3, 3), let x = 0. False statement.

For (3, 4), let x = 3.5. True statement.

For (4, infinity), let x = 6. False statement.

I understand why the solution is [-4, -3] U [3, 4].

So, any value of x in the intervals [-4, -3] and [3, 4] including -4, -3, 3, and 4 satisfy the given inequality. I may post two more quadratic inequality to get additional practice.