
Question
2x^27X>15
Let me just break it down till the end.
(x5)(2x+3)>0
x50 2x+3=0
x=5 x=3/2
When I do my test point and let it be 0 It looks something like
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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?

Re: Question
Go back to (x5)(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

Re: Question
Actually :
$\displaystyle 2\left(\frac{3}{2}\right)^27\left(\frac{3}{2}\right)15=0$