Q54 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
If the roots of the equation, $8 x^3+6 p x^2+3 q x-27=0$ are in a geometric progression, then $q^2+9 p^2+6 p q+q / p=$
- Correct answerA-3
- B-10
- C6
- D0
Solution
Let roots of the equation
$8 x^3+6 p x^2+3 q x-27=0 \text { are } \frac{a}{r}, a, \text { ar. }$
$\therefore \quad \frac{a}{r}+a+a r=\frac{-6 p}{8}=\frac{-3 p}{4}$
$\Rightarrow \quad a\left(\frac{1}{r}+1+r\right)=\frac{-3 p}{4}$ $\ldots$ (i)
$\left(\frac{a}{r} \cdot a\right)+(a \cdot a r)+\left(\frac{a}{r} \cdot a r\right)=\frac{3 q}{8}$
$\Rightarrow \quad a^2\left(\frac{1}{r}+r+1\right)=\frac{3 q}{8}$ $\ldots$ (ii)
and $\quad \frac{a}{r} \times a \times a r=\frac{27}{8}$
$\Rightarrow \quad a^3=\frac{27}{8} \Rightarrow a=\frac{3}{2}$ $\ldots$ (iii)
From Eqs. (i) and (ii), we get
$a=\frac{3 q / 8}{-3 p / 4}=-\frac{q}{2 p}$ $\ldots$ (iv)
From Eqs. (iii)and (iv), we have
$\frac{-q}{2 p}=\frac{3}{2} \Rightarrow-q=3 p$
Now, $\quad q^2+9 p^2+6 p q+\frac{q}{p}$
$\Rightarrow \quad 9 p^2+9 p^2+6 p(-3 p)+\frac{(-3 p)}{p}$
$\Rightarrow \quad 18 p^2-18 p^2-3=-3$
Hence, the correct answer is option: 'A'.