Let f(x)=3ax2−4bx+c(a,b,c∈R,a≠0) where a,b,c are in A.P. Then the equation f(x)=0 has
Since a,b,c are in A.P., so,
2b=a+c
4b2=(a+c)2
The discriminant of the given functionf(x)=3ax2−4bx+c is,
D=16b2−12ac
=4(a+c)2−12ac
=4[(a2+c2+2ac)−3ac]
=4(a2+c2−ac)
=4(a2+c2−2ac+ac)
=4((a−c)2+ac)
Case 1: If a andc are of opposite signs, then, D=(+)ve.
Case 2: If a andc are of same signs, then, D=(+)ve.
This shows that f(x)=0 has two unequal real roots.