Given that $ f(x)= ax^2+bx+c$, where $b = a + 1$ and $c=b+1$. One zero of the function is $x=6$. Find the other zero of $f$.
I actually have the answer to the question but only a brief explanation of some of the steps. If I define $R$ to be the missing root the question is calling for. I understand that $6R = \frac{a+2}{a}$, but why is $6+R=\frac{-(a+1)}{a}$?
When I attempted this problem by myself I thought $6+R = \frac{a+1}{a}$?
I am sure I am missing something very fundamental any explanation would be appreciated.
Hint: $$(x-x_1)(x-x_2)=0$$ $$x^2-(x_1+x_2)x+(x_1.x_2)=0$$