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

Results 1 to 3 of 3

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

- Joined
- Mar 2012
- From
- LA
- Posts
- 68

## 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?

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

- Joined
- Mar 2012
- From
- Sheffield England
- Posts
- 440
- Thanks
- 76

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

- Joined
- Nov 2011
- From
- Crna Gora
- Posts
- 420
- Thanks
- 64