Asked by April
Solve for x.
x/(x-2) + (2x)/[4-(x^2)] = 5/(x+2)
Please show work!
x/(x-2) + (2x)/[4-(x^2)] = 5/(x+2)
Please show work!
Answers
Answered by
Steve
putting 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
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
Answered by
MathMate
x/(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
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
Answered by
MathMate
x=(5±(√15)*i)/2
If you do not expect complex roots, please check for typos in the original question.
If you do not expect complex roots, please check for typos in the original question.
Answered by
April
Thank 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!
Answered by
MathMate
You're welcome!
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