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.