edited by
699 views

1 Answer

Best answer
1 votes
1 votes

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

selected by
Answer:

Related questions

0 votes
0 votes
1 answer
1
go_editor asked May 5, 2016
920 views
The number of roots common between $x^3 + 3 x^2 + 4x + 5 = 0$ and $x^3 + 2 x^2 + 7x +3 =0$ is$0$$1$$2$$3$
0 votes
0 votes
0 answers
2
go_editor asked Mar 29, 2016
276 views
If the equation $x^3 – ax^2 + bx – a = 0$ has three real roots, then it must be the case that$b=1$$b \neq 1$$a=1$$a \neq 1$
0 votes
0 votes
1 answer
3
go_editor asked Apr 30, 2016
654 views
Let $f(x) = ax^2 + bx +c$, where $a, b$ and $c$ are certain constants and $a \neq 0$. It is known that $f(5) = -3 f(2)$ and that $3$ is a root of $f(x)=0$.What is the val...