Q53 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics

If the roots of $x^3+a x^2+b x+c=0$ are in arithmetic progression with common difference 1 , then

  • Correct answer
    A
    $9 c=a(b-2)$
  • B
    $9 c=a(2-b)$
  • C
    $9 c-a^2(b-2)=0$
  • D
    $9 c-a^2(2-b)=0$

Solution

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 $\left(-\frac{a}{3}\right)\left[\frac{b+1}{3}-1\right]=-c \Rightarrow \frac{a}{3}\left(\frac{b-2}{3}\right)=c$ $\Rightarrow \quad 9 c=a(b-2)$

Hence, the correct answer is option: 'A'.

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