Solve 2/[(x^2)-x] = (1/x) + 4/(x^2-1)

May 2010
2
0
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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dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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)
Are the terms in the red part of exponent or not?
 
May 2010
2
0
2/((x^2)-x) = (1/x) + (4/((x^2)-1)

Hopefully that makes it easier to understand. The exponents are in red.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
\(\displaystyle \frac{2}{x^2}-x=\frac{1}{x}+\frac{4}{x^2}-1\)

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.
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
2/((x^2)-x) = (1/x) + (4/((x^2)-1)

Hopefully that makes it easier to understand. The exponents are in red.
\(\displaystyle \frac{2}{x^2-x} = \frac{1}{x} + \frac{4}{x^2-1}
\)

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

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
 
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