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# If the roots of the equation x^3 – ax^2 + bx – c = 0 - Roots -PS

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If the roots of the equation x^3 – ax^2 + bx – c = 0 - Roots -PS  [#permalink]

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15 Aug 2018, 06:59
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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

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If the roots of the equation x^3 – ax^2 + bx – c = 0 - Roots -PS  [#permalink]

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15 Aug 2018, 07:36
1
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 $$b= -1$$
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Re: If the roots of the equation x^3 – ax^2 + bx – c = 0 - Roots -PS  [#permalink]

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15 Aug 2018, 07:40
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

It's a cubic equation..
Let the roots be w,y,z
1) sum of all roots =w+y+z= a
2) sum of two roots at one time =wy+wz+yz=b
3) product of roots=wyz=c
So all 3 consecutive numbers as roots will give sum as the least when one is 0..
The other two -1 and 1, so roots are -1,0,1
And b=-1*1+0*1+0*-1=-1

B
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: If the roots of the equation x^3 – ax^2 + bx – c = 0 - Roots -PS &nbs [#permalink] 15 Aug 2018, 07:40
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# If the roots of the equation x^3 – ax^2 + bx – c = 0 - Roots -PS

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