The following MCQ(single correct option) is from JEE Main 2022
Let $S$ be the set of all integral values of $\alpha$ for which the sum of squares of two real roots of the quadratic equation $3x^2+(\alpha-6)x+(\alpha+3)=0$ is minimum. Then $S$:
- Is an empty Set
- Is a singleton
- Contains exactly two elements
- Contains more than two elements
This solution uses the condition for the equation to have non-complex roots: $$\text{Discriminant}\geq0\implies\alpha(\alpha-24)\geq0\implies(\alpha\leq0)\cup(\alpha\geq24)$$
If $a$ and $b$ are roots of the quadratic, the sum of squares using Vieta's formulas for the sum and product of roots: $$a^2+b^2=(a+b)^2-2ab=\frac{\alpha^2-18\alpha+18}{9}=f(\alpha)$$
The above quadratic is minimized at its vertex, $\alpha=\frac{-b}{2a}=9$, which is rejected since for this value the roots are not real.
My Question: The function to be minimized is a quadratic with restricted domain, and hence it has a minimum at the left endpoint(see graph). Hence, the given solution is incorrect.
However, two real roots (as opposed to a real root, or repeated roots), means the roots must be distinct, which means the discrimant must be $>$, and not $\geq$. If you redo the solution using this constraint, the endpoints of the domain are not included, and hence there is no minimum for values.
Hence, the solution arrives at the right answer, but needs to be corrected as above. Is my thought process correct?