If a, b, c are real number such that a+b+c=0 then the quadratic equation 3ax^2 +2bx+c=0 has.
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Step-by-step explanation:
Let f
′
(x)=3ax
2
+2bx+c
⟹f(x)=ax
3
+bx
2
+cx+d (on integration)
Rolle's theorem states that if f(x) be continuous on [a,b], differentiable on (a,b) and f(a)=f(b) then there exists some c between a and b such that f
′
(c)=0
let (a,b)=(0,1)
f(0)=d and f(1)=a+b+c+d
But, f(0)=f(1).
⟹d=a+b+c+d
⟹a+b+c=0
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