The given equation is
$$x^2-x+a-3<0$$
And
$$x^2 + x -(a+3) <0$$
The right answer is 6, which can be obtained by replacing the inequity with an equality, making the discriminant positive and adjusting for a negative root to obtain the interval of $a$
$$x^2-x+a-3=0$$
And
$$x^2+x-(a+3)=0$$
So $a>-\frac{13}{4}$ and $a< \frac{13}{4}$
Also $$x=\frac{1\pm \sqrt{13-4a}}{2}$$
So $$\sqrt{13-4a}>1$$
$$a<3$$
That gives $a\in (-\frac{13}{4}, 3)$ ie. 6 integral values
But this method doesn’t sit right with me, since values satisfying the inequality were supposed to be found, not the equality. Maybe it’s just a technicality, but I am not convinced. Can I get an explanation for this?
I know a similar question already exists in the site, but I found it unsatisfactory and it doesn’t answer my specific problem
Best Answer
If $a=3$ (orange line) then we have equality and $x=0$ which is a right boundary case. Hence we need $a<3$.
If $\displaystyle a=-\frac{13}{4}$ (violet line) then we have the left boundary where equality holds for $x<0$. So, we have $\displaystyle a>-\frac{13}{4}$ for inequality.