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If the roots of $x^3+a x^2+b x+c=0$ are in arithmetic progression with common difference 1 , then
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The correct answer is:
$9 c=a(b-2)$
Let the roots of $x^3+a x^2+b x+c=0$ are $\alpha-1, \alpha, \alpha+1$ as roots are in AP with common difference 1 .
Now, sum of roots $=3 \alpha=-a$ $\ldots(\mathrm{i})$
Sum of product of roots taking two at a time
$=(\alpha-1) \alpha+(\alpha+1) \alpha+(\alpha-1)(\alpha+1)=b$
$\Rightarrow 3 \alpha^2-1=b$
and product of roots $=\alpha\left(\alpha^2-1\right)=-c$
From Eqs. (i), (ii) and (iii), we have
$\left(-\frac{a}{3}\right)\left[\frac{b+1}{3}-1\right]=-c \Rightarrow \frac{a}{3}\left(\frac{b-2}{3}\right)=c$
$\Rightarrow \quad 9 c=a(b-2)$
Now, sum of roots $=3 \alpha=-a$ $\ldots(\mathrm{i})$
Sum of product of roots taking two at a time
$=(\alpha-1) \alpha+(\alpha+1) \alpha+(\alpha-1)(\alpha+1)=b$
$\Rightarrow 3 \alpha^2-1=b$
and product of roots $=\alpha\left(\alpha^2-1\right)=-c$
From Eqs. (i), (ii) and (iii), we have
$\left(-\frac{a}{3}\right)\left[\frac{b+1}{3}-1\right]=-c \Rightarrow \frac{a}{3}\left(\frac{b-2}{3}\right)=c$
$\Rightarrow \quad 9 c=a(b-2)$
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