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why is (x-2)(x+1) > 0 = x>2 and X<1 ?
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maybe this will help you
I can't open that Myin.
just draw the function and you will see thats what i have in the pdf
When I see it I get it. But I wouldn't be able to understand the algebra...
(x-2)(x+1) > 0 x-2>0, x+1>0 x>2 x>-1
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that's what I said nick but it's wrong. If x is greater than -1 you get a negative y value.
since it is of the form (x-a)(x-b)>0 then the roots will be x<a or x>b hence it is x<-1 or x>2
For the function to be >0 (a positive number) either: each bracket term must be +ve : x>2 AND x>-1 == x>2 or: each bracket term must be -ve: x<2 AND x<-1 == x<-1 so for f(x)>0: x must be <-1 or >2 but not anywhere between those two
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