How do you solve the rational equation #x/(x-2) + (2x)/(4-x^2) = 5/(x+2)#?

1 Answer
Feb 13, 2016

Answer:

#x=(5+-sqrt(15)i)/2#

Explanation:

#x/(x-2)+(2x)/(4-x^2)=5/(x+2)#

First put all the denominatores is standard for by multiplying #(2x)/(4-x^2)#by (-1/-1)

#x/(x-2)-(2x)/(x^2-4)=5/(x+2)#

#x^2-4=(x+2)(x-2)#, so we can easily put the fractions with the same denominator:

#(x(x+2))/(x^2-4)-(2x)/(x^2-4)=(5(x-2))/(x^2-4)#

No we can eliminate the denominators. The equivalent of multiplying both sides of equation by #x^2-4#, but we must keep in mind that x!=+-2, because these values woul put #x^2-4=0#

#x(x+2)-(2x)=5(x-2)#

#x^2+2x-2x=5x-10#

#x^2-5x+10=0#

Now we must use the expression for solving a second degree polynom:

#x=(5+-sqrt(5^2-4*10))/2#

#x=(5+-sqrt(-15))/2#

#x=(5+-sqrt(15)i)/2#