Quadratics: Sum of Solutions

1) Multiply by denominator of right expression: $\displaystyle(x-2)(x+4){x\over{x-2}} = (x-2)(x+4){6x\over{(x-2)(x+4)}}$

2) Simplify: $(x+4)x=6x$

3) Expand brackets and move terms to one side: $x^2 + 4x - 6x = 0$ , $x^2 - 2x = 0$

4) Factor: $x(x-2)=0$

5) Possible solutions are $2$ and $0$ .

6) Remove extraneous solutions. Since $x = 2$ (one of the possible solutions), and one side of the equation's denominator is $x-2$ , substituting would give $2 - 2=0$ (cannot divide by 0), meaning that $x = 2$ cannot be a solution to the equation (it is an extraneous solution). Applying $x=0$ to the original equation gives: ${0\over{-2}}={0\over{-8}}$ , $0=0$ (solution verified).

7) Our remaining solution is $x = 0$ , therefore the sum of solutions is $0$ .