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

1 Answer
Steve M
Aug 25, 2017

There are three solutions:

# x=0, +- sqrt(5/2) #

Explanation:

We have:

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

# :. 1/(x+1)+1/(x-1)+1/(x+2)+1/(x-2) = 0 #

# :. ( (x-1)+(x+1) ) / ( (x+1)(x-1) ) + ( (x-2)+(x+2) ) / ( (x+2)(x-2) ) = 0 #

# :. ( 2x ) / ( x^2-1 ) + ( 2x ) / ( x^2-4 ) = 0 #

# :. 2x{ ( 1 ) / ( x^2-1 ) + ( 1 ) / ( x^2-4 ) } = 0 #

# :. x{ ( (x^2-4) + (x^2-1) ) / ( (x^2-1)(x^2-4) ) }= 0 #

# :. x{ ( 2x^2-5 ) / ( (x^2-1)(x^2-4) ) }= 0 #

# :. x ( 2x^2-5 ) = 0 #

Thus:

# x=0 #
# 2x^2-5 =0 => x = +- sqrt(5/2) #

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