Any Math gurus?

[SuperGroove]

Hi all. I'm having a little problems here as I study for a test that was announced yesterday. My teacher says that we'll be having the test because he won't be there. Great...it'd be nice if I could actually ask HIM for help...but teacher's...ya know...they're not for teaching. They're only there for a part time job while they go earn their master's or something to that effect. ARGH!

Anyways...seems like everyone is out tonight, and the ones I can call...well they's like me. Stupid. SO where's what I'm working on:

Vertical and Horizontal Asymtotes and finding domains as well as solving for the infamous x variable. I don't know how to do some characters, so bear with me.

1. the cube root of (x^2 +3x+1)=2 ; Translation: x squared plus 3x plus 1 is equal to 2. Any ideas? This problem is solve for X

2. (x^2 +2x-3)^(1/3)=5 ; Essentially that's "the cube root of x squared plus 2x minus 3 is equal to 5. Solve for X

Out of the 50 problems I did, these two gave me the most difficult time. What do I need to do?[/b]



Also, how do I tell what functions do? i.e, if X--->infinity, then what does F(x) do?

Thanks,
Paul

 

[Napalm381]

What do you need to find on these specifically? Solve for x? Graphs?

 

Aren't these just simple quadratic equations? Cube each side to get rid of the cube root and stuff'em in your quadratic formula. No asymptotes, domain -inf -> +inf.

 

[SuperGroove]

for both, I just need to solve for X

Paul

 

[Napalm381]

Like Bobtist said, just cube each side so you have a simple quadratic equation (ax^2+bx+c=k), rearrange and plug it into the quadratic formula.

 

[BiggieN]

uh...no math guru esta aqui.

 

I don't think either of those two have real solutions, either. Have fun.

 

[Napalm381]

Bobtist- they appear to have real solutions to me- the determinant is greater than zero. Did I miss something?

 

[eakers]

1. bring the 2 over and factor
x^2+3x-1=0
use the quadratic eqn -b+or-[(b^2-4ac)^1/2]/2ac
oh and there is answers cuz the negaive c will make sure you have a positve under the square root

2. again bring the 5 over you have to add it to -3*1/3=1 which =6 but then common factor of 1/3 makes it -2
1/3(x^2 +2x-2)=0
then quadratic.

thats what i think anyways.

 

lol, yep! Somehow I decided I was powerful enough to break the rules of mathematics and turn x=125 into x+125=0. Oy, that won't go over well on the final tomorrow.

Thanks

 

[SuperGroove]

For the first one I got:

negative 3 plus or minus the square root of 37. This is all divided by two.

For the second problem I got:

x^2+2x-3=125; x squared plus two, minus 3, equals 125

I haven't put it into the quadratic formula yet. Am I good to go? No?

 

Correct.

 

[Theslowone]

first part of one is (-3+ sqrt(13))/2
2nd part of oen is (-3 - sqrt(13))/2

first part of two is (-2+sqrt(516))/2
second " " " (-2-sqrt(516))/2

I think

 

[SuperGroove]

Holy mother of... Bobtist, you gotta be kidding me. Could it be...that I actually got a math problem correct? Pats myself on back...time to go relax NFS: PU. That's what I'll do. I've got this sweet arse 3.6 Turbo waiting to be driven. It spins it's wheels all the way through 3rd gear. I'm wondering if I can get a heavy duty differential in the game....maybe two!

Thanks!
Paul

 

[ManSnake]

I have a friend, his friend's friend's friend is pretty good at math!

 

[xtreme2k]

Also, how do I tell what functions do? i.e, if X--->infinity, then what does F(x) do?

when x tends to infinity,
both f(x) also tends to infinity...

 

[smartt]

When you get through with that easy one try this. If a green train leaves the station at 9:30 traveling west and a blue train leaves the station at 10:30 traveling east, who will be the next president? Damn algebra!