If the roots of x^3+a x^2+b x+c=0 are in arit | vedclass

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(A) Let the roots of $x^3+a x^2+b x+c=0$ be $\alpha-1, \alpha, \alpha+1$ as the roots are in $AP$ with common difference $1$.Sum of roots $= (\alpha-1

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$9 c=a(b-2)$

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(A) Let the roots of $x^3+a x^2+b x+c=0$ be $\alpha-1, \alpha, \alpha+1$ as the roots are in $AP$ with common difference $1$.Sum of roots $= (\alpha-1) + \alpha + (\alpha+1) = 3\alpha = -a \Rightarrow \alpha = -\frac{a}{3} \quad \dots(i)$Sum of product of roots taken two at a time $= (\alpha-1)\alpha + \alpha(\alpha+1) + (\alpha-1)(\alpha+1) = b$$\Rightarrow \alpha^2 - \alpha + \alpha^2 + \alpha + \alpha^2 - 1 = b$ $\Rightarrow 3\alpha^2 - 1 = b \quad \dots(ii)$Product of roots $= (\alpha-1)\alpha(\alpha+1) = \alpha(\alpha^2-1) = -c \quad \dots(iii)$Substitute $\alpha = -\frac{a}{3}$ into $(ii)$: $3(-\frac{a}{3})^2 - 1 = b$ $\Rightarrow \frac{a^2}{3} - 1 = b$ $\Rightarrow a^2 = 3(b+1)$.Substitute $\alpha = -\frac{a}{3}$ into $(iii)$: $(-\frac{a}{3})((-\frac{a}{3})^2 - 1) = -c$$\Rightarrow \frac{a}{3}(\frac{a^2}{9} - 1) = c$ $\Rightarrow \frac{a}{3}(\frac{3(b+1)}{9} - 1) = c$$\Rightarrow \frac{a}{3}(\frac{b+1}{3} - 1) = c$ $\Rightarrow \frac{a}{3}(\frac{b-2}{3}) = c$ $\Rightarrow 9c = a(b-2)$.

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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