CAT 2008 | Question: 14

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. $\frac{1}{\sqrt{3} }$

Answer 1

option 2

This can be solved by sum and product of roots of cubic equation.

$Ax^3 +Bx^2 + Cx +D=0$ where pqr are three roots

so p*q*r = -D/A

p+q+r = -B/A

pq+qr+pr = C/A

for the given eqn $x^{3} - ax^{2} + bx - c=0$ consecutive roots are (k-1), k and (k+1)

So, (k-1)* k *(k+1) = +c

(k-1) + k + (k+1)=+a

k(k-1) + k(k+1) + (k-1)(k+1)=+b

=> 3k^2 -1 = b (here k will be alwys +ve)

b's minimum value is possible only when k is 0

and b= 3(0) - 1= - 1