Question

If the 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. \(-\frac{1}{\sqrt{3}}\)
• B. \(-1\)
• C. 0
• D. 1
• E. \(\frac{1}{\sqrt{3}}\)

Hint:

When roots are in arithmetic progression, symmetry helps simplify sum and product calculations using Vieta’s formulas.

Answer 1

The Correct Option is B

Let the roots be \(m-1, m, m+1\). By Vieta's formulas: \[ a = (m-1) + m + (m+1) = 3m \] \[ b = (m-1)m + m(m+1) + (m+1)(m-1) \] Simplifying: \[ b = (m^2 - m) + (m^2 + m) + (m^2 - 1) = 3m^2 - 1 \] To minimize \(b\), minimize \(m^2\). The smallest possible \(m^2\) is \(0\) when \(m = 0\), giving: \[ b = -1 \] Thus, the smallest possible value is \(\boxed{-1}\).