If the roots of the equation $x^{3} - a x^{2} + b x - c = 0$ are three consecutive integers, then what is the smallest possible value of b?

- \frac{1}{\sqrt{3}}

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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
$(n - 1) n + n (n + 1) + (n + 1) (n - 1) = b$
$\Rightarrow n^{2} - n + n^{2} + n + n^{2} - 1 = b$
$\Rightarrow 3 n^{2} - 1 = b$
The value of b will be minimum when the value of $n^{2}$ is minimum i.e., $n^{2} = 0$
Hence, minimum value of b = -1.

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If the roots of the equation x3−ax2+bx−c=0 are three consecutive integers, then what is the smallest possible value of b?

The correct option is B -1Let roots are (n - 1), n and (n + 1) Sum of products of roots taken two at a time = b (n−1)n+n(n+1)+(n+1)(n−1)=b ⇒n2−n+n2+n+n2−1=b ⇒3n2−1=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.

If the roots of the equation $x^{3} - a x^{2} + b x - c = 0$ are three consecutive integers, then what is the smallest possible value of b?