Go back to (x-5)(2x+3)>0

This is true if both brackets are positive or both brackets negative.

Both positive if x>5 and both negative if x<-3/2

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- Apr 6th 2012, 05:57 PM #1

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## Question

2x^2-7X>15

Let me just break it down till the end.

(x-5)(2x+3)>0

x-5-0 2x+3=0

x=5 x=-3/2

When I do my test point and let it be 0 It looks something like

<--//////-----------------------------------------------------///////--------->

True -3 over 2 the middle is F and 5 is also true and the interval notation is (-∞.-3/2)U(5,∞).

My question is, how can -3/2 be greater than 0?

- Apr 6th 2012, 10:40 PM #2

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- Apr 6th 2012, 10:49 PM #3

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