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Need some Assistance. (x3)^2/(x+1)(x2)>0
I'm supposed to solve it and express my answer in interval notation.
 9 months ago
 9 months ago
Need some Assistance. (x3)^2/(x+1)(x2)>0 I'm supposed to solve it and express my answer in interval notation.
 9 months ago
 9 months ago

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RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
dw:1372837027496:dw
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
is this \[\frac{(x3)^{2}}{(x+1)(x2)}\]?
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
Yes. that is the equation.
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
ok so the top is >0 for all x does not equal 3 right?
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
so we need to make sure the bottom is > 0 right?
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
dw:1372837247856:dw
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
i have forgot 3
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
the bottom is a parabola that opens up, what are its x intercepts?
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
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?
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
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
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
(infinity,1)U(2,3)U(3,infinity)
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
I have another quick question
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
I had another problem like this and I solved it and got x<1. How would I write that in interval notation?
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
if it was <= (infinity,1]
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
The original problem was \[3(1\frac{ 3 }{4 }x)>5\frac{ 1 }{ 4 }x\]
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
3(9/4)x>5(1/4)x (8/4)x>53 2x<2 x<1 correct
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
\[x <1\] I wrote the interval notation like so (infinity,1). But for some reason, my professor marked off points and circled interval notation.
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
you sure it didn't say set builder?
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
The question said Solve the inequalities and express your answer in Interval notation.
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
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 <=
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
just go talk to your teacher, I make mistakes grading all the time....
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
\[\frac{x=1 }{ 4}\ge4\]
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
is that plus or minus?
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
I know your supposed to get two answers. I got one which was \[x \ge15\]
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
\[\frac{ x+1 }{4 }\]
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
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 b<x<b when you have a>b we get a<b and a>b
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
the very last and should be or
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
*when you have a>b we get a<b or a>b
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
Okay, So if I want the other answer I set the 4 to a negative answer?
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
so you have abs((x+1)/4)>=4 so (x+1)/4>=4 gives the solutions you already have
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
now we do (x+1)/4<=4 x+1<=16 x<=17
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
set it negative and flip the sign
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
Ah, that's what confused me.. Didn't know what to do with my sign
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
you could have done (x+1)/4<=4 but as a rule I just make the "non absolute value side" negative
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
Good rule! With interval notation, how would I write this?
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
I think its \[(\infty,17]U(\infty,15]\]
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
(infinity,17]U[15,infinity)
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
think of the number line, write it in the order it would fall on the number line
 9 months ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
and note that (infinity,5] does not make since, because if x were to be in there then infinity < x < 5
 9 months ago

Brutus.Best ResponseYou've already chosen the best response.0
But, Thank you so much for the help.
 9 months ago
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