Algebra-8 (CAT-2008)

If the roots of the equation x³-ax²+bx-c=0 are three consecutive integers, then what is the smallest possible value of b?

(A)-1/√3        (B)-1       (C)0      (4)1      (5)1/√3

Solution follows here:

Solution:

This problem is a typical simplification type.

x³-ax²+bx-c=0 ---(1)

Let the three routes be n-1, n, n+1:  Hence these values satisfy the equation (1).

(n-1)³-a(n-1)²+b(n-1)-c=0                    ---(2)

n³-an²+bn-c=0                                      ---(3)

(n+1)³-a(n+1)²+b(n+1)-c=0                 ---(4)          

(3) – (2) yields => n³-(n-1)³-a(n²-(n-1)²)+b(n-(n-1)) = 0

=> 1-3n+3n² -a(2n-1)+b = 0               ---(5)

(4) – (3) yields => (n+1)³-n³-a((n+1)²-n²)+b(n+1-n) = 0

=> 1+3n+3n² -a(2n+1)+b = 0             ---(6)

Multiply (5) with (2n+1) and (6) with (2n-1):

(2n+1)(1-3n+3n²) - a(2n+1) (2n-1) + b(2n+1) = 0 ---(7)

(2n-1)( 1+3n+3n²) - a(2n+1) (2n-1) + b(2n-1) = 0 ---(8)

(8) – (7) yields:

6n²-2+b(-2) = 0 => -2b = 2-6n²  => b = 3n²-1

n² is always positive, and its minimum value is zero.

Hence minimum value of b = 3(0)-1 = -1

Answer (B)