Harshgmat wrote:

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) - 1/√3

(B) - 1

(C) 0

(D) 1

(E) 1/√3

OA:B

General cubic equation with roots \(p,q\) and \(r\) will be given by \((x-p)(x-q)(x-r)=0\).

\((x-p)(x-q)(x-r) =(x^2-(p+q)x+pq)(x-r)=x^3 -(p+q)x^2 +pqx -rx^2 +r(p+q)x-pqr)=x^3-(p+q+r)x^2+(pq+qr+rp)x-pqr\)

Equating above expression to \(x^3 – ax^2 + bx – c\), we get

\(b= (pq+qr+rp)\)

let \(3\) consecutive number be \(n-1,n,n+1\)

\(b = (n-1)n+n(n+1)+(n-1)(n+1) = 3n^2-1\)

\(n^2≥0\), minimum value of \(n^2\) is \(0\), leading to minimum value of \(b= -1\)

_________________

Good, good Let the kudos flow through you