SOLUTION: x+4/x^3-3x^2-4x+12 + 3/x^2-5x+6 = 3x/x^2-4. This problem is not working out for me, I come up with x^2+3x+8=0

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: x+4/x^3-3x^2-4x+12 + 3/x^2-5x+6 = 3x/x^2-4. This problem is not working out for me, I come up with x^2+3x+8=0      Log On


   

Question 75687: x+4/x^3-3x^2-4x+12 + 3/x^2-5x+6 = 3x/x^2-4.
This problem is not working out for me, I come up with x^2+3x+8=0
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
I assume you mean: (a few brackets would help)
:
%28%28x%2B4%29%29%2F%28%28x%5E3-3x%5E2-4x%2B12%29%29 + 3%2F%28%28x%5E2-5x%2B6%29%29 = %283x%29%2F%28%28x%5E2-4%29%29
:
All the denominators will factor:
%28%28x%2B4%29%29%2F%28%28x-2%29%28x%2B2%29%28x-3%29%29%29 + 3%2F%28%28x-3%29%28x-2%29%29%29 = %283x%29%2F%28%28x-2%29%28x%2B2%29%29
:
A common denominator is (x-2)(x+2)(x-3), multiply each term by this and you have:
:
(x+4) + 3(x+2) = 3x(x-3)
:
x + 4 + 3x + 6 = 3x^2 - 9x
:
4x + 10 = 3x^2 - 9x
:
Arrange as a quadratic equation:
3x^2 - 9x - 4x - 10 = 0
3x^2 - 13x - 10 = 0
:
This factors to:
(3x + 2)(x -5) = 0
:
3x = -2
x = -2/3
and
x = +5
:
I substituted 5 for x in the original equation and got equality,
I'll let you check out the x = -2/3 solution