2x^2 + 5x - 3 >= 0

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ryana

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Jan 14, 2008
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Question:
How and why does the inequality sign in the answer get flipped for x = -3?

Problem: Solve the inequality
2x^2 + 5x - 3 >= 0

Factored:
(x + 5/4)^2 = 49/16

Eliminate exponent:
x + 5/4 = +- 7/4

Equation solved:
x = -3 OR x = 1/2

My answer: Replace the = with >=
x >= -3 OR x >= 1/2

Correct answer:
x <= -3 OR x >= 1/2
 
My approach would be to find the critical points.
2x^2+5x-3=0
(2x-1)(x+3)=0
Critical points are x=1/2 and x=-3.
Next, I would examine what happens when x < -3. I would find that the function is positive.
Then I would examine what happens if -3 < x < 1/2. I would find that the function is negative.
Finally by examining what happens when x>1/2, I would find that the function is positive.
This tells me that 2x^2+5x-3>0 when x<-3 and when x>1/2.
I don't see where any "flipping" was involved.
 
Could you please explain your factoring technique and method of isolating the critical points

My book only lists 2 methods for factoring Trinomials.
1. Factor using coefficients if the leading coefficient = 1
2. Factor by decomposition if leading coefficient > 1

Since no two numbers added together to = 5 and multiply to equal -3 or -6 I'm lost.
 
ryana said:
Could you please explain your factoring technique and method of isolating the critical points

My book only lists 2 methods for factoring Trinomials.
1. Factor using coefficients if the leading coefficient = 1
2. Factor by decomposition if leading coefficient > 1

Since no two numbers added together to = 5 and multiply to equal -3 or -6 I'm lost.
Click to expand...

(2) * (-3) = -6 = (6) * (-1)

and

(6) + (-1) = 5
 
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