Given quadratic equation,
3x2+(α−6)x+(α+3)=0Let,
a and
b are the roots of the equation,
∴a+b=−and
ab=For real roots,
D≥0⇒(α−6)2−4⋅3⋅(α+9)≥0⇒α2−12α+36−12α−36≥0⇒α2−24α≥0⇒α(α−24)≥0
∴ Real roots of the equation possible for
α>24 or
α<0.
Now, sum of square of roots
=a2+b2=(a+b)2−2ab=−2⋅===f(x)∴ Sum of square of roots are minimum when
a2+b2= minimum.
∴f(α)min=Value of quadratic equation
α2−18α+18 is minimum at
α=−=−=9But for real roots
α should be less than 0 or greater than 24 .
So, there is no value of
α in the range
α>24∪α<0 where sum of squares of two real roots is minimum.
∴S is an empty set.