Evaluating Chosen Numbers in Quadratic Inequality

[mathdad]

Solve the inequality.

2x^2 + 7x + 5 > 0

Factor LHS.

2x^2 + 2x + 5x + 5 > 0

2x(x + 1) + 5(x + 1) > 0

(2x + 5)(x + 1) > 0

2x + 5 = 0

2x = -5

x = -5/2

x + 1 = 0

x = -1

Plot x = -5/2 and x = -1 on a number line.

<--------(-5/2)----------(-1)----------->

Pick a number from each interval.

Let x = -4 for (-infinity, -5/2).

Let x = -1/2 for (-5/2, -1).

Let x = 0 for (-1, infinity).

Do I avaluate the chosen numbers per interval in the original question or the factored form?

 

[Euge]

Hi RTCNTC,

Plug in your test points into the expression $(2x + 5)(x + 1)$ to determine the signs in the respective intervals. For example, let's consider your test point $x = -4$. Plug in $-4$ for $x$ in the expression $(2x + 5)(x + 1)$ to get $(-7)(-3) = 21$, which has positive sign. Hence, $(2x + 5)(x + 1) > 0$ in the interval $(-\infty, -5/2)$. Now try the other two test points and see what you get.

 

[mathdad]

Ok. I will continue as you say tomorrow. Going to work now. Math keeps me from drowning in my despair. Thank God for math.

 

[mathdad]

Pick a number from each interval.

Let x = -4 for (-infinity, -5/2).

Let x = -1/2 for (-5/2, -1).

Let x = 0 for (-1, infinity).

For x = -4, we get true.

For x = -1/2, we get true.

For x = 0, we get true.

We exclude -5/2 and -1.

Solution: (-infinity, -5/2) U (-1, infinity)

Correct?

 

[Euge]

Yes, that's correct.

 

[mathdad]

Very good. What about the other quadratic inequality question? Is that one correct as well?

 

[MarkFL]

In this problem, you have a function representing a parabola that opens upwards. We should then expect, given that it has two real roots, that the intervals on which is is positive will be on either side of the two roots, rather than in between. :D

 

[mathdad]

Thank you everyone. Two more inequality questions later tonight, and yes, I will show my work.