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Question:
Roots of the equation x^3 - ax^2 + bx - c = 0 are three consecutive integers then what is the smallest possible value of b?
A) -1/root3
B) -1
C) 0
D) 1
E) 1/root 3
Solution:
a-1, a and a+1
Min value of b = Min value of (a-1)a + a(a+1) + (a+1)(a-1)
= Min value of a^2 - a + a^2 + a + a^2 - 1
= Min value of 3a^2 - 1
= -1
y-1, y, y+1
b = (y-1)*y+(y-1)(y+1)+(y+1)*y = 3y^2-1.. minimum value -1
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