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
Solve for x.
x/(x-2) + (2x)/[4-(x^2)] = 5/(x+2)
Please show work!
5 answers
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
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.
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!
You're welcome!