Can somebody help me solve this equation? I have gotten to a certain point then get stuck. Thank you!

2/x^2-x = (1/x) + (4/x^2-1)

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- May 11th 2010, 04:30 PMconnroy17Solve 2/[(x^2)-x] = (1/x) + 4/(x^2-1)
Can somebody help me solve this equation? I have gotten to a certain point then get stuck. Thank you!

2/x^2-x = (1/x) + (4/x^2-1) - May 11th 2010, 04:52 PMdwsmith
- May 11th 2010, 05:33 PMconnroy17
2/((x^2)-x) = (1/x) + (4/((x^2)-1)

Hopefully that makes it easier to understand. The exponents are in red. - May 11th 2010, 05:45 PMdwsmith
$\displaystyle \frac{2}{x^2}-x=\frac{1}{x}+\frac{4}{x^2}-1$

Move to one side and simplify:

$\displaystyle \frac{2}{x^2}+x+\frac{1}{x}-1=0$

Obtain same common denominator.

You can do that one. - May 11th 2010, 06:01 PMskeeter
$\displaystyle \frac{2}{x^2-x} = \frac{1}{x} + \frac{4}{x^2-1}

$

$\displaystyle \frac{2}{x(x-1)} = \frac{1}{x} + \frac{4}{(x+1)(x-1)}$

common denominator is $\displaystyle x(x+1)(x-1)$ ...

$\displaystyle \frac{2(x+1)}{x(x-1)(x+1)} = \frac{1(x+1)(x-1)}{x(x+1)(x-1)} + \frac{4x}{x(x+1)(x-1)}$

note that x cannot equal 0 , 1 , or -1 (why?)

with a common denominator, the numerators form the equation ...

$\displaystyle 2(x+1) = (x+1)(x-1) + 4x$

solve this quadratic for x