How do you solve #(x^2-x-6)/(x+2)+(x^3+x^2)/x=3#?

1 Answer
Jan 17, 2017

#x = -1+-sqrt(7)#

Explanation:

#3 = (x^2-x-6)/(x+2) + (x^3+x^2)/x#

#color(white)(3) = ((x-3)color(red)(cancel(color(black)((x+2)))))/color(red)(cancel(color(black)((x+2)))) + (color(red)(cancel(color(black)(x)))(x^2+x))/color(red)(cancel(color(black)(x)))#

#color(white)(3) = x-3+x^2+x = x^2+2x-3#

Subtract #3# from both ends to get:

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

#color(white)(0) = x^2+2x+1-7#

#color(white)(0) = (x+1)^2-(sqrt(7))^2#

#color(white)(0) = ((x+1)-sqrt(7))((x+1)+sqrt(7))#

#color(white)(0) = (x+1-sqrt(7))(x+1+sqrt(7))#

Hence:

#x = -1+-sqrt(7)#

These are both solutions of the original equation, since neither value causes any denominator to be #0#.