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 \[3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0\] is minimum. Then S :
Solutions · 1
Official solution
Given quadratic equation,
\[3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0\]
Let, a and b are the roots of the equation,
\[\therefore\] \[a + b = - {{\alpha - 6} \over 3}\]
and \[ab = {{\alpha + 3} \over 3}\]
For real roots,
\[D \ge 0\]
\[ \Rightarrow {(\alpha - 6)^2} - 4\,.\,3\,.\,(\alpha + 9) \ge 0\]
\[ \Rightarrow {\alpha ^2} - 12\alpha + 36 - 12\alpha - 36 \ge 0\]
\[ \Rightarrow {\alpha ^2} - 24\alpha \ge 0\]
\[ \Rightarrow \alpha (\alpha - 24) \ge 0\]
\[\therefore\] \[\alpha > 24\] or \[\alpha < 0\]
\[\therefore\] Real roots of the equation possible for \[\alpha > 24\] or \[\alpha < 0\].
Now, sum of square of roots
=a2+b2=(a+b)2−2ab
\[ = {{{{(\alpha - 6)}^2}} \over 9} - 2\,.\,{{(\alpha + 3)} \over 3}\]
\[ = {{{\alpha ^2} - 12\alpha + 36 - 6\alpha - 18} \over 9}\]
\[ = {{{\alpha ^2} - 18\alpha + 18} \over 9} = f(x)\]
\[\therefore\] Sum of square of roots are minimum when a2+b2= minimum.
\[\therefore\] \[f{(\alpha )_{\min }} = {{{\alpha ^2} - 18\alpha + 18} \over 9}\]
Value of quadratic equation \[{\alpha ^2} - 18\alpha + 18\] is minimum at \[\alpha = - {b \over {2a}} = - {{( - 18)} \over {2\,.\,1}} = 9\]
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.