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Need some Assistance. (x-3)^2/(x+1)(x-2)>0 I'm supposed to solve it and express my answer in interval notation.

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is this \[\frac{(x-3)^{2}}{(x+1)(x-2)}\]?
Yes. that is the equation.

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Other answers:

ok so the top is >0 for all x does not equal 3 right?
so we need to make sure the bottom is > 0 right?
i have forgot 3
the bottom is a parabola that opens up, what are its x intercepts?
-1,2 Right?
so the bottom is greater than 0 when x<-1 and x>2, because it opens up and crosses y=0 at -1,2 with me?
So far yes.
so we know the top is grater than 0 when x is not 3 we know the bottom is greater than 0 when x<-1 and x>2 so the whole thing is greater than 0 when x<-1 and x>2 and x does not equal 3
Okay, I get it now.
good deal:)
I have another quick question
I had another problem like this and I solved it and got x<-1. How would I write that in interval notation?
if it was <= (-infinity,-1]
The original problem was \[3(1-\frac{ 3 }{4 }x)>5-\frac{ 1 }{ 4 }x\]
3-(9/4)x>5-(1/4)x -(8/4)x>5-3 2x<-2 x<-1 correct
\[x <-1\] I wrote the interval notation like so (-infinity,-1). But for some reason, my professor marked off points and circled interval notation.
you sure it didn't say set builder?
The question said Solve the inequalities and express your answer in Interval notation.
he/she is wrong look at non ending interval note: we have ) not ] because we have < not <=
just go talk to your teacher, I make mistakes grading all the time....
Okay, thank you!
One more equation?
\[|\frac{x=1 }{ 4}|\ge4\]
is that plus or minus?
I know your supposed to get two answers. I got one which was \[x \ge15\]
Plus. My bad
\[|\frac{ x+1 }{4 }|\]
ok so this is strange and you may need to think about it a bit, but when you have |a|=b we get a=b and a=-b when you have |a|b we get a-b
the very last and should be or
*when you have |a|>b we get a-b
Okay, So if I want the other answer I set the 4 to a negative answer?
so you have abs((x+1)/4)>=4 so (x+1)/4>=4 gives the solutions you already have
not exactly
now we do (x+1)/4<=-4 x+1<=-16 x<=-17
set it negative and flip the sign
Ah, that's what confused me.. Didn't know what to do with my sign
you could have done -(x+1)/4<=4 but as a rule I just make the "non absolute value side" negative
Good rule! With interval notation, how would I write this?
I think its \[(-\infty,-17]U(\infty,15]\]
think of the number line, write it in the order it would fall on the number line
Oh, I see.
and note that (infinity,5] does not make since, because if x were to be in there then infinity < x < 5
You lost me there..
But, Thank you so much for the help.

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