How do you solve #\frac { 3} { x + 1} + \frac { 3} { x + 2} + \frac { 3} { x - 1} + \frac { 3} { x - 2} = 0#?

1 Answer
George C.
Aug 8, 2018

#x=0#, #x=1/2sqrt(10)# or #x=-1/2sqrt(10)#

Explanation:

Given:

#3/(x+1)+3/(x+2)+3/(x-1)+3/(x-2) = 0#

By observation #x=0# is a solution, but knowing that does not really help us immediate find any other solutions.

Let's sort out some common denominators and simplify the given expression...

#0 = 3/(x+1)+3/(x+2)+3/(x-1)+3/(x-2)#

#color(white)(0) = 3/(x+1)+3/(x-1)+3/(x+2)+3/(x-2)#

#color(white)(0) = (3(x-1))/((x-1)(x+1))+(3(x+1))/((x-1)(x+1))+(3(x-2))/((x-2)(x+2))+(3(x+2))/((x-2)(x+2))#

#color(white)(0) = (3(x-1)+3(x+1))/(x^2-1)+(3(x-2)+3(x+2))/(x^2-4)#

#color(white)(0) = 6x(1/(x^2-1)+1/(x^2-4))#

#color(white)(0) = 6x((x^2-4)+(x^2-1))/((x^2-1)(x^2-4))#

#color(white)(0) = 6x((2x^2-5)/((x^2-1)(x^2-4)))#

#color(white)(0) = 6x((2x^2-5)/((x^2-1)(x^2-4)))#

#color(white)(0) = 3x((4x^2-10)/((x^2-1)(x^2-4)))#

#color(white)(0) = 3x(((2x)^2-(sqrt(10))^2)/((x^2-1)(x^2-4)))#

#color(white)(0) = 3x(((2x-sqrt(10))(2x+sqrt(10)))/((x^2-1)(x^2-4)))#

Hence #x=0#, #x=1/2sqrt(10)# or #x=-1/2sqrt(10)#

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