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Mathematics 21 Online
OpenStudy (anonymous):

why is (x-2)(x+1) > 0 = x>2 and X<1 ?

myininaya (myininaya):

maybe this will help you

OpenStudy (anonymous):

I can't open that Myin.

myininaya (myininaya):

just draw the function and you will see thats what i have in the pdf

OpenStudy (anonymous):

When I see it I get it. But I wouldn't be able to understand the algebra...

OpenStudy (anonymous):

(x-2)(x+1) > 0 x-2>0, x+1>0 x>2 x>-1

OpenStudy (anonymous):

that's what I said nick but it's wrong. If x is greater than -1 you get a negative y value.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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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