## Wednesday, 26 October 2011

### 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