anonymous
  • anonymous
Need some Assistance. (x-3)^2/(x+1)(x-2)>0 I'm supposed to solve it and express my answer in interval notation.
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1372837027496:dw|
zzr0ck3r
  • zzr0ck3r
is this \[\frac{(x-3)^{2}}{(x+1)(x-2)}\]?
anonymous
  • anonymous
Yes. that is the equation.

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zzr0ck3r
  • zzr0ck3r
ok so the top is >0 for all x does not equal 3 right?
anonymous
  • anonymous
Right.
zzr0ck3r
  • zzr0ck3r
so we need to make sure the bottom is > 0 right?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
|dw:1372837247856:dw|
anonymous
  • anonymous
i have forgot 3
zzr0ck3r
  • zzr0ck3r
the bottom is a parabola that opens up, what are its x intercepts?
anonymous
  • anonymous
-1,2 Right?
zzr0ck3r
  • zzr0ck3r
right
zzr0ck3r
  • zzr0ck3r
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?
anonymous
  • anonymous
So far yes.
zzr0ck3r
  • zzr0ck3r
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
zzr0ck3r
  • zzr0ck3r
(-infinity,-1)U(2,3)U(3,infinity)
anonymous
  • anonymous
Okay, I get it now.
zzr0ck3r
  • zzr0ck3r
good deal:)
anonymous
  • anonymous
I have another quick question
zzr0ck3r
  • zzr0ck3r
oki
anonymous
  • anonymous
I had another problem like this and I solved it and got x<-1. How would I write that in interval notation?
zzr0ck3r
  • zzr0ck3r
(-infinity,-1)
zzr0ck3r
  • zzr0ck3r
if it was <= (-infinity,-1]
anonymous
  • anonymous
The original problem was \[3(1-\frac{ 3 }{4 }x)>5-\frac{ 1 }{ 4 }x\]
zzr0ck3r
  • zzr0ck3r
3-(9/4)x>5-(1/4)x -(8/4)x>5-3 2x<-2 x<-1 correct
anonymous
  • anonymous
\[x <-1\] I wrote the interval notation like so (-infinity,-1). But for some reason, my professor marked off points and circled interval notation.
zzr0ck3r
  • zzr0ck3r
you sure it didn't say set builder?
anonymous
  • anonymous
The question said Solve the inequalities and express your answer in Interval notation.
zzr0ck3r
  • zzr0ck3r
he/she is wrong http://www.regentsprep.org/Regents/math/ALGEBRA/AP1/IntervalNot.htm look at non ending interval note: we have ) not ] because we have < not <=
zzr0ck3r
  • zzr0ck3r
just go talk to your teacher, I make mistakes grading all the time....
anonymous
  • anonymous
Okay, thank you!
zzr0ck3r
  • zzr0ck3r
np
anonymous
  • anonymous
One more equation?
anonymous
  • anonymous
Inequality*
zzr0ck3r
  • zzr0ck3r
sure
anonymous
  • anonymous
\[|\frac{x=1 }{ 4}|\ge4\]
zzr0ck3r
  • zzr0ck3r
is that plus or minus?
anonymous
  • anonymous
I know your supposed to get two answers. I got one which was \[x \ge15\]
anonymous
  • anonymous
Plus. My bad
anonymous
  • anonymous
\[|\frac{ x+1 }{4 }|\]
zzr0ck3r
  • zzr0ck3r
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
zzr0ck3r
  • zzr0ck3r
the very last and should be or
zzr0ck3r
  • zzr0ck3r
*when you have |a|>b we get a-b
anonymous
  • anonymous
Okay, So if I want the other answer I set the 4 to a negative answer?
zzr0ck3r
  • zzr0ck3r
so you have abs((x+1)/4)>=4 so (x+1)/4>=4 gives the solutions you already have
zzr0ck3r
  • zzr0ck3r
not exactly
zzr0ck3r
  • zzr0ck3r
now we do (x+1)/4<=-4 x+1<=-16 x<=-17
zzr0ck3r
  • zzr0ck3r
set it negative and flip the sign
anonymous
  • anonymous
Ah, that's what confused me.. Didn't know what to do with my sign
zzr0ck3r
  • zzr0ck3r
you could have done -(x+1)/4<=4 but as a rule I just make the "non absolute value side" negative
anonymous
  • anonymous
Good rule! With interval notation, how would I write this?
anonymous
  • anonymous
I think its \[(-\infty,-17]U(\infty,15]\]
zzr0ck3r
  • zzr0ck3r
(-infinity,-17]U[15,infinity)
zzr0ck3r
  • zzr0ck3r
think of the number line, write it in the order it would fall on the number line
anonymous
  • anonymous
Oh, I see.
zzr0ck3r
  • zzr0ck3r
and note that (infinity,5] does not make since, because if x were to be in there then infinity < x < 5
zzr0ck3r
  • zzr0ck3r
<=5
anonymous
  • anonymous
You lost me there..
anonymous
  • anonymous
But, Thank you so much for the help.

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