IIT JEE , AIEEE Forum - Mathematics , Algebra

sachin_gupta1991

Q. If a+b+c=0, then prove that the quadratic equation 3ax²+2bx+c=0 has

at least one root in [0,1].

nishantsingh89

f(x)= 3ax^2 + 2bx + c

on integrating f(x) we get

F(x)= ax^3 + bx^2 + c + d
(d integration constant)
put x=0

F(x) = d

put x=1
F(x) = a+b+c +d = d { a+b+c =0}

now F(x) being a polynomial function is continuous in [0,1] and differentiable in (0,1)
and F(0)=F(1)
so by rolle's theorem
there exists atleast one point c { 0<c<1}
such that
F'(c) =0 { F'(x) = f(x) }
it implies that there is atleast one root of the equation 3ax^2 + 2bx + c i interval [0,1]

nudge me if you have doubts...............

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