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Question

If the roots of the equation x3ax2+bxc=0 are three consecutive integers, then what is the smallest possible value of b?


A
13
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B
-1
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C
1
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Solution

The correct option is B -1
Let roots are (n - 1), n and (n + 1)
Sum of products of roots taken two at a time = b
(n1)n+n(n+1)+(n+1)(n1)=b
n2n+n2+n+n21=b
3n21=b
The value of b will be minimum when the value of n2 is minimum i.e., n2=0
Hence, minimum value of b = -1.

Quantitative Aptitude

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