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

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Answer 1 · 2018-01-12T10:58:23

Solution: #x = 2.5 +- 1.94i (2dp) #

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

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

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

by #(x+2)(x-2)# on both sides we get ,

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

#x^2+cancel(2x)-cancel(2x) =5x-10# or

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

#x^2-5x + (5/2)^2 = 25/4-10# or

#(x-2.5)^2 = -3.75 or (x-2.5) = +-sqrt (-3.75) # or

#(x-2.5) = +-sqrt ((3.75)i^2) # or

#x = 2.5+-sqrt ((3.75)i^2); i^2=-1 # or

#x = 2.5 +- 1.94i(2dp) #

Solution: #x = 2.5 +- 1.94i (2dp) # [Ans]