How do you solve #2/(x(x+2))+3/x=4/(x-2)#?

1 Answer
MeneerNask
Nov 26, 2017

We can solve this by multiplying everything by the least common multiple #x(x+2)(x-2)#

Explanation:

Note: We have to reject the solutions #x=0,x=2,x=-2#, as they would make one or more of the fractions undetermined.

The factors not in the denominators will not be canceled and end up in the numerators:
#->(2*cancelx(cancel(x+2))(x-2))/(cancelx(cancel(x+2)))+(3*cancelx(x+2)(x-2))/cancelx=(4*x(x+2)(cancel(x-2)))/cancel(x-2)#

#->2(x-2)+3(x+2)(x-2)=4x(x+2)#

Work out the parentheses and put everything to one side:
#->2x-4+3x^2-12=4x^2+8x#

#->-x^2-6x-16 =0->x^2+6x+16=0#

The Discriminant #D=b^2-4ac=6^2-4*1*16=-28<0#

Since #D<0#, there are no real solutions.

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