How do you solve #-3 + \frac{1}{x+1}=\frac{2}{x}# by finding the least common multiple?

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Answer 1 · 2018-07-24T13:42:53

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Given that

#-3+1/{x+1}=2/x#

#-3+1/{x+1}-2/x=0#

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

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

#{-3x^2-4x-2}/{x(x+1)}=0#

#-3x^2-4x-2=0\quad (\forall \ x\ne 0, x\ne -1)#

#3x^2+4x+2=0#

#B^2-4AC=4^2-4(3)(2)=-8<0#

The given equation has no real root .

The complex roots are given by quadratic formula as follows

#x=\frac{-4\pm\sqrt{(4)^2-4(3)(2)}}{2(3)}#

#=\frac{-4\pm2i\sqrt2}{6}#

#=\frac{-2\pmi\sqrt2}{3}#