The sum of all parameter $a\in\mathbb{R}$ values for which the graphs of the functions $y=(a+2)x^2-ax-3$ and $y=ax-4$ have exactly one common point is? The answer in textbook is -1.
This is what I tried:
I need a common point so I equate the two functions. $$(a+2)x^2-ax-3=ax-4 \to (a+2)x^2-2ax+1=0$$ I need only one common point, therefore only one solution, so the discriminant is zero. $$D=0 \to 4a^2-4(a+2)=0 \to 4a^2-4a-8=0 \to a^2-a-2=0$$ Using quadratic formula for $a^2-a-2=0$, I get solutions $a_1=2 \land a_2=-1$ whose sum is 1.
I checked the numbers too many times, I guess my logic is bad.
As per AnuragA's comment, one should consider the possibility of $a+2=0$.