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15 Aug 2018, 07:59
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
67% (01:42) correct
33%
(00:12)
wrong
based on 3
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) - 1/√3
(B) - 1
(C) 0
(D) 1
(E) 1/√3
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15 Aug 2018, 08:36
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Harshgmat
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 \(b= -1\)
15 Aug 2018, 08:40
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Harshgmat
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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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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