assuming x≠0, multiply by x^2 to get
14 - 5x = x^2
x^2+5x-14 = 0
(x+7)(x-2) = 0
I think you can take it from there, no?
Solve algebraically:
14/x^2 - 5/x = 1
Thank you!!! (:
4 answers
That's exactly what I was thinking, then I was also thinking, wouldn't this work?
14 - 5/x = x^2
14 - 5 = x^3
9 = x^3
???
14 - 5/x = x^2
14 - 5 = x^3
9 = x^3
???
no, you have to multiply every term by x^2. This method skips the 5/x term. Then your second line skips the 14 term when multiplying by x.
What kind of algebra is that?
What kind of algebra is that?
...I don't know that's exactly why I was asking actually. cool. thanks.