Can't remember how to do this: (3/(x^2))+(5/x)-12=0

[amyp]

Hi there. I'm trying to help a friend with her math, but it's been a few years and I cannot remember how to solve these problems. (And I don't remember what it's called to look at examples and figure out how to do it)

Here are the problems she is supposed to solve:

(3/(x^2)) + (5/x) - 12=0

And...

(3/(x-1)) + (10/(x+1)) = 4

With this one I tried making a common denominator and then added like terms in the numerator but got lost trying to solve for x after that...

Thank you so much for the help!

 

[DrMike]

(3/(x^2)) + (5/x) - 12=0

If you let y=1/x, you get 3y[sup:37szxi7o]2[/sup:37szxi7o]+5y-12=0. Do you know how to solve it after that?

(3/(x-1)) + (10/(x+1)) = 4

With this one I tried making a common denominator and then added like terms in the numerator but got lost trying to solve for x after that...

Thank you so much for the help!

You're on the right track - what did you get just before you got stuck?

 

[Deleted member 4993]

Duplicate post:

viewtopic.php?f=9&t=34081

These are all variations of quadratic equations.