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

- anonymous

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- anonymous

|dw:1372837027496:dw|

- zzr0ck3r

is this \[\frac{(x-3)^{2}}{(x+1)(x-2)}\]?

- anonymous

Yes. that is the equation.

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## More answers

- zzr0ck3r

ok so the top is >0 for all x does not equal 3 right?

- anonymous

Right.

- zzr0ck3r

so we need to make sure the bottom is > 0 right?

- anonymous

Yes.

- anonymous

|dw:1372837247856:dw|

- anonymous

i have forgot 3

- zzr0ck3r

the bottom is a parabola that opens up, what are its x intercepts?

- anonymous

-1,2 Right?

- zzr0ck3r

right

- 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

So far yes.

- 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

(-infinity,-1)U(2,3)U(3,infinity)

- anonymous

Okay, I get it now.

- zzr0ck3r

good deal:)

- anonymous

I have another quick question

- zzr0ck3r

oki

- anonymous

I had another problem like this and I solved it and got x<-1. How would I write that in interval notation?

- zzr0ck3r

(-infinity,-1)

- zzr0ck3r

if it was <=
(-infinity,-1]

- anonymous

The original problem was
\[3(1-\frac{ 3 }{4 }x)>5-\frac{ 1 }{ 4 }x\]

- zzr0ck3r

3-(9/4)x>5-(1/4)x
-(8/4)x>5-3
2x<-2
x<-1
correct

- 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

you sure it didn't say set builder?

- anonymous

The question said Solve the inequalities and express your answer in Interval notation.

- 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

just go talk to your teacher, I make mistakes grading all the time....

- anonymous

Okay, thank you!

- zzr0ck3r

np

- anonymous

One more equation?

- anonymous

Inequality*

- zzr0ck3r

sure

- anonymous

\[|\frac{x=1 }{ 4}|\ge4\]

- zzr0ck3r

is that plus or minus?

- anonymous

I know your supposed to get two answers. I got one which was \[x \ge15\]

- anonymous

Plus. My bad

- anonymous

\[|\frac{ x+1 }{4 }|\]

- 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

the very last and should be or

- zzr0ck3r

*when you have |a|>b we get a

**-b**

**
**- anonymous

Okay, So if I want the other answer I set the 4 to a negative answer?

- zzr0ck3r

so you have abs((x+1)/4)>=4
so
(x+1)/4>=4 gives the solutions you already have

- zzr0ck3r

not exactly

- zzr0ck3r

now we do
(x+1)/4<=-4
x+1<=-16
x<=-17

- zzr0ck3r

set it negative and flip the sign

- anonymous

Ah, that's what confused me.. Didn't know what to do with my sign

- zzr0ck3r

you could have done -(x+1)/4<=4
but as a rule I just make the "non absolute value side" negative

- anonymous

Good rule! With interval notation, how would I write this?

- anonymous

I think its
\[(-\infty,-17]U(\infty,15]\]

- zzr0ck3r

(-infinity,-17]U[15,infinity)

- zzr0ck3r

think of the number line, write it in the order it would fall on the number line

- anonymous

Oh, I see.

- zzr0ck3r

and note that (infinity,5] does not make since, because if x were to be in there then
infinity < x < 5

- zzr0ck3r

<=5

- anonymous

You lost me there..

- anonymous

But, Thank you so much for the help.

**
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