Q83 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics

If the roots of $x^3-p x^2+q x-r=0$ are in AP. Then,

  • Correct answer
    A
    $2 p^3-9 p q+27 r=0$
  • B
    $2 p^3+9 p q-27 r=0$
  • C
    $2 p^3-8 p q+27 r=0$
  • D
    $2 p^3-9 p q+28 r=0$

Solution

Let $a-d, a, a+d$ are roots of $$ x^3-p x^2+q x-r=0 $$ Sum of roots $=-\frac{b}{a}$ $$ \begin{aligned} a-d+a+a+d & =-\frac{(-p)}{1} \\ 3 a & =p \\ a & =\frac{p}{3} \end{aligned} $$ Since, $a=\frac{P}{3}$ should be satisfied by given equation. So, put $x=\frac{p}{3}$ in Eq. (i) $$ \left(\frac{p}{3}\right)^3-P\left(\frac{p}{3}\right)^2+q\left(\frac{p}{3}\right)-r=0 $$ $$ \begin{aligned} \frac{p^3}{27}-\frac{p^3}{9}+\frac{p q}{3}-r & =0 \\ p^3-3 p^3+9 p q-27 r & =0 \\ -2 p^3+9 p q-27 r & =0 \\ 2 p^3-9 p q+27 r & =0 \end{aligned} $$ Hence, option (1) is correct.

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

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