How do I solve a quadratic inequality using factoring?

[mathdad]

Solve the inequality.

x^2 + 4x - 32 < 0

Factor LHS.

(x - 4) (x + 8) < 0

x - 4 = 0

x = 4

x + 8 = 0

x = -8

Plot x = 4 and x = -8 on a number line.

<--------(-8)----------(4)----------->

Pick a number from each interval.

Let x = -10 for (-infinity, -8).

Let x = 0 for (-8, 4).

Let x = 6 for (4, infinity).

Do I avaluate the chosen numbers per interval in the original question or the factored form (x - 4) (x + 8) < 0?

 

[MarkFL]

You can look at the signs of each factor to determine the sign of the quadratic in a given interval containing a text value, and so determine if that interval is part of the solution. You really only need to evaluate one interval, and then use the fact that for roots of odd multiplicity, the original expression will change sign across that root, and for roots of even multiplicity, the expression won't change sign.

All of the roots in this problem are of odd multiplicity, so you know the sign of the expression will alternate across all intervals. :D

 

[mathdad]

So, it is negative across all three intervals. I will continue later tonight or tomorrow. Going to work now.

 

[MarkFL]

So, it is negative across all three intervals...

How did you make that determination?

 

[mathdad]

I made a typo. The signs alternate across all three intervals. I will complete BOTH inequality questions tomorrow. Look for 5 questions (not math questions) through PM in 15 minutes.

 

[mathdad]

Pick a number from each interval.

Let x = -10 for (-infinity, -8).

Let x = 0 for (-8, 4).

Let x = 6 for (4, infinity).

For x = -10, we get False.

For x = 0, we get True.

For x = 6, we get False.

We exclude the end points.

The solution to the original inequality is found in (-8, 4).

Correct?

 

[MarkFL]

Yes, we have a parabola opening upwards, and given that it has two real roots, we should expect to find it to be negative in between its roots. :D

 

[mathdad]

Cool. Two more math questions later tonight.