Question

Solve each of the following equations and also verify your solution :

(x + 2)(x+3)+(x-3)(x-2)-2x(x+1)=0

Answer 1

$(x+2)(x+3)+(x-3)(x-2)-2x(x+1)=0\Rightarrow [x^2+(2+3)x+2\times 3]+[x^2+(-3-2)x+(-3\left)\right(-2\left)\right]-2x^2-2x=0\Rightarrow x^2+5x+6+x^2-5x+6-2x^2-2x=0\Rightarrow x^2+x^2-2x^2+5x-5x-2x+6+6=0=-2x+12=0$
Substracting 12 from both sides,
$-2x+12-12=0-12\Rightarrow -2x=-12$
Dividing by - 2,
$\frac{-2x}{-2}=\frac{-12}{-2}\Rightarrow x=6\therefore x=6$
Verification :
$L.H.S.=(x+2)(x+3)+(x-3)(x-2)-2x(x+1)=(6+2)(6+3)+(6-3)(6-2)-2\times 6(6+1)=8\times 9+3\times 4-12\times 7=72+12-84=84-84=0=R.H.S.$