Quadratic Equation and Inequalities Part 2

Given quadratic equation,$3 x^2+(\alpha-6) x+(\alpha+3)=0$Let, $a$ and $b$ are the roots of the equation,$\therefore a+b=-\; \frac{\alpha-6}{3}$and $ab=\; \frac{\alpha+3}{3}$For real roots,$D \ge 0$$\Rightarrow (\alpha-6)^2-4 \cdot 3 \cdot (\alpha+9) \ge 0$$\Rightarrow \alpha^2-12\alpha+36-12\alpha-36 \ge 0$$\Rightarrow \alpha^2-24\alpha \ge 0$$\Rightarrow \alpha(\alpha-24) \ge 0$$\therefore$ Real roots of the equation possible for $\alpha > 24$ or $\alpha < 0$.Now, sum of square of roots$= a^2+b^2$$= (a+b)^2 - 2ab$$=\; \frac{(\alpha-6)^2}{9} - 2 \cdot \; \frac{\alpha+3}{3}$$=\; \frac{\alpha^2 -12\alpha +36 -6\alpha -18}{9}$$=\; \frac{\alpha^2 -18\alpha +18}{9} = f(x)$$\therefore$ Sum of square of roots are minimum when $a^2+b^2=$ minimum.$\therefore f(\alpha)_{\min} = \; \frac{\alpha^2 -18\alpha +18}{9}$Value of quadratic equation $\alpha^2 -18\alpha +18$ is minimum at $\alpha = -\; \frac{b}{2a} = -\; \frac{-18}{2.1} = 9$But for real roots $\alpha$ should be less than 0 or greater than 24 .So, there is no value of $\alpha$ in the range $\alpha > 24 \cup \alpha < 0$ where sum of squares of two real roots is minimum.$\therefore S$ is an empty set.