solving rational equations

[mathhelper]

this appears to be straight forward just wanted to run it past a few people.
1/x -1/x+2 = 1/58
now I solve for LCD 58x(x+2)???

which gives me 1(x+2)58 -1(58x)=-x(x+2) over 58x(x+2)???

Now I mulitply both sides by 58x(x+2) which leaves:

58(x+2)-58x= -x(x+2) which becomes
58x+116 -58x= -x^2-2x set it equal to zero by adding x^2 and 2 to the other side.

x^2 -2x +116= 0

Now I am stuck... Can someone help me figure out where I went wrong

 

[Deleted member 4993]

this appears to be straight forward just wanted to run it past a few people.
1/x -1/x+2 = 1/58
now I solve for LCD 58x(x+2)???

which gives me 1(x+2)58 -1(58x)=-x(x+2) over 58x(x+2)???

Now I mulitply both sides by 58x(x+2) which leaves:

58(x+2)-58x= -x(x+2) which becomes
58x+116 -58x= -x^2-2x set it equal to zero by adding x^2 and 2 to the other side.

x^2 -2x +116= 0

Now I am stuck... Can someone help me figure out where I went wrong

1/x - 1/(x+2) = 1/58

multiply both sides by 58*x*(x+2)

58(x+2) - 58x = x(x+2)

x[sup:3so89vwy]2[/sup:3so89vwy] + 2x - 116 = 0

\(\displaystyle x \ \ = \frac{-2 \pm \sqrt {4 + 4\cdot 116}}{2}\)

\(\displaystyle x \ \ = {-1 \pm \sqrt {234}\)

and that's it....

 

[BigGlenntheHeavy]

\(\displaystyle \frac{1}{x}-\frac{1}{x+2} \ = \ \frac{1}{58}\)

\(\displaystyle \frac{x+2-x}{x(x+2)} \ = \ \frac{1}{58}\)

\(\displaystyle \frac{2}{x^2+2x} \ = \ \frac{1}{58}\)

\(\displaystyle x^2+2x \ = \ 116\)

\(\displaystyle x^2+2x-116 \ = \ 0\)

\(\displaystyle x \ = \ \frac{-2\pm\sqrt{4+4(116)}}{2}\)

\(\displaystyle x \ = \ \frac{-2\pm6\sqrt{13}}{2}\)

\(\displaystyle x \ = \ -1 \ \pm3\sqrt{13}\)

 

[mathhelper]

Did my work appear to be correct up to the point I became stuck??

If the #'s changed to
1/x (1 over x) - 1/(x+1) (1 over x+1) =1/56 (1 over 56) does my denominator become 56x(x+1) or 56(x+1)