For $x>a$
$$9-x^2 >x-a$$
$$x^2 + x -(9+a)<0$$
So
$$1+4(9+a)<0$$
$$a<\frac{-37}{4}$$
Similarly for $x<a$
$$x^2-x+a-9<0$$
So $a>\frac{37}{4}$
Now in both cases, we need at least one $x<0$
In case one, $a<0$ and $x>a$ so $x$ can be both positive and negative, so I don’t know how to deal with that
Similar problem in case 2.
Another inequality I found was $a>-9$, obtained by solving the quadratic for first case, giving $x\in (\frac{1-\sqrt {37+4a}}{2}, \frac{1+\sqrt {37+4a}}{2})$, and then setting the lower limit to $<0$
I can also find another inequality from the second case, but there are way too many inequalities to choose from and the condition that $x>a$ and $x<a$ is throwing me off.
Also is a downward parabola even possible for these cases? The equation demands it to be downward if $f(x)<0$ for all $x$, but the coefficient of $x^2$ is positive here.
Best Answer
ok so as you have said:
Case 1: $x>a$
$$9-x^2 >x-a$$ $$x^2 + x -(9+a)<0$$ Now, $x^2 + x -(9+a)$ is an upward facing parabola and we need to find the region of where it is below the x axis; because after all, that's where it is less than zero. For an upward facing parabola to have a region below the x-axis it must have two real roots, and hence the discriminant must be greater than zero. $\implies b^2-4ac> 0 \implies 1+4(9+a)>0 \implies a>\dfrac{-37}{4}$. So when $a>\dfrac{-37}{4}$ that upward facing parabola has two real roots and hence there exists a region below the x-axis. But this is not the only thing we need. We need this region below the x-axis to have at least one negative value of x. Notice the x coordinate of the vertex of this parabola = $ \dfrac{-b}{2a}= \dfrac{-1}{2}$, which is less than zero. That means at least one of the two roots is negative, and at least one negative value exists in the solution set of: $x^2 + x -(9+a)<0$ iff $a>\dfrac{-37}{4}$.
But we also said $x$ is strictly greater than $a$ initially before removing the absolute value. This means for at least one negative solution, $a$ must be strictly less than zero, because if $a$ exceeds zero, then $x$ must also and then there will be no negative solutions for x... So for $x>a, a\in (\dfrac{-37}{4},0).$
Case 2: $x<a$
$$9-x^2 >-(x-a)$$ $$x^2 - x +(a-9)<0$$
Again discriminant $D>0 \implies a<\dfrac{37}{4}$, but here vertex of parabola is positive = 0.5. This means at least one root is positive. To make sure that the other root is negative we need the parabola to be below the $x$ axis when $x=0.$ So $0^2 - 0 +(a-9)<0 \implies a<9.$ We also need to make sure this negative root is less than $a$ because our initial constraint was $x<a.$ So when does the line x =a, meet the parabola at the negative root? $a^2 - a +(a-9)=0 \implies a = -3 \implies a\in [-3,9)$
So $a \in (\dfrac{-37}{4},0) \cup [-3,9] \implies a \in (\dfrac{-37}{4},9)$