Here is the problem statement:
According to the parabola in the figure of the function $f(x) = ax^2 + bx + c$, it is given that $a + b + c = 0$. The points $A$ and $B$ are the x-intersects, and the point $C$ is the y-intersect of the parabola. Let the origin be the point $O$, and determine the area of the triangle $COA$.
Here are my solving steps and why I think the plot doesn't seem right.
- From the figure, it is apparent that $C = 3$, which leaves me with $a + b = -3$, from which I can write $b = - a - 3$.
- I find the roots of the parabola, which turn out to be $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Substituting $b = - a - 3$ and $c = 3$ the roots are $x_1 = 1$ and $x_2 = \frac{3}{a}$.
However, since the value from under the square root on step 2 above is a positive value, I can assume that $x_1$ > $x_2$, simply because I used $+$ of the $\pm$ for $x_1$, and the $-$ for $x_2$. But the problem's answer accepts 1 to be the x coordinate of the point $A$ and proceeds on calculating the area from there with $\frac{3\times1}{2}=1.5$. I'm intending to appeal for the problem, but can't make sure whether or not I'm missing something.

In solving the quadratic equation, x= $\frac{(a+3)\pm\sqrt{(a+3)^2-12a}}{2a}$ =$\frac{(a+3)\pm|a-3|}{2a}$
when a$\geq$3, $x_1$=1>$x_2$=$\frac{3}{a}$;x2 is the smaller root. In this case, the area of $\Delta$ AOC has many values pending on value of a.
when a$\leq$3, $x_1$=1<$x_2$=$\frac{3}{a}$;1 is the small root. In this case, the area of $\Delta$ AOC is $\frac{3}{2}$.
Hope this clarify the picture.