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
Si...
Discussion between Sid and Aditya 
Discussion between Sid and Aditya