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October 30, 2014

October 30, 2014

Posted by **April** on Friday, July 12, 2013 at 2:49pm.

x/(x-2) + (2x)/[4-(x^2)] = 5/(x+2)

Please show work!

- Math -
**Steve**, Friday, July 12, 2013 at 3:01pmputting everything over a common denominator of (x-2)(x+2), we have

x^2 = 5(x-2)

x^2 - 5x + 10 = 0

then just solve that for x

Make sure that x ≠ ±2 since those values are not allowed in the original equation

- Math -
**MathMate**, Friday, July 12, 2013 at 3:08pmx/(x-2) + (2x)/[4-(x^2)] = 5/(x+2)

This calls for finding the common denominator, and then solving for x using the numerators.

Do not forget to exclude the asymptotes values of x where the denominator becomes zero.

LCM (lowest ommon multiple) of denominators (x-2), (4-x²) and (x+2) is (x²-4), since (x-2)(x+2)=(4-x²)

Provided that x≠2 and x≠-2, then we can write the equation above as:

[x(x+2)-2x]/(x²-4) = 5(x-2)/(x²-4)

provided x≠2 and x≠-2

Equating numerators,

x²+2x-2x=5(x-2)

=>

x²-5x+10=0

However, the solution does not have real roots. The complex roots are:

x=(5±√*i)/2

- Math -
**MathMate**, Friday, July 12, 2013 at 3:09pmx=(5±(√15)*i)/2

If you do not expect complex roots, please check for typos in the original question.

- Math -
**April**, Friday, July 12, 2013 at 3:19pmThank you guys! That is what I got. I wasn't expecting the complex roots.. That's why I wanted to check whether or not I was doing it right!

- Math :) -
**MathMate**, Friday, July 12, 2013 at 4:31pmYou're welcome!

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