Q182 / 270 · NDA 2024 (Phase 1) · Mathematics
If $a, b$ and $c(a \gt 0, c \gt 0)$ are in GP, then consider the following in respect of the equation $a x^2+b x+c=0:$
1. The equation has imaginary roots.
2. The ratio of the roots of the equation is $1: \omega$ where $\omega$ is a cube root of unity.
3. The product of roots of the equation is $\left(\frac{b^2}{a^2}\right)$.
Which of the statements given above are correct?
- A1 and 2 only
- B2 and 3 only
- C1 and 3 only
- Correct answerD1,2 and 3
Solution
Given that $a, b, c$ are in GP
$\therefore \quad b^2=a c$
Now, $a x^2+b x+c=0$
$\begin{aligned}
\mathrm{D} & =b^2-4 a c \\
& =a c-4 a c=-3 a c \lt 0
\end{aligned}$
1. $\therefore$ Roots are imaginary (correct)
Let, $\quad a=2, b=4, c=8$
$\begin{aligned}
\therefore \quad x^2+2 x+4 & =0 \\
\Rightarrow \quad x & =\frac{-2 \pm \sqrt{4-16}}{2}=-1 \pm \sqrt{3}
\end{aligned}$
2. Ratio of roots $=\frac{-1+\sqrt{3}}{-1-\sqrt{3}}=\frac{\omega}{\omega^2}=\frac{1}{\omega}$ (correct)
3.$\begin{aligned}
\text { Product of roots } & =\frac{(-1+\sqrt{3})(-1-\sqrt{3})}{2} \times 4 \\
& =\omega \cdot \omega^2 \cdot 4=4=\frac{b^2}{a^2}
\end{aligned}$ (correct)
Hence, statements 1,2 and 3 are correct.
Hence, the correct answer is option: 'D'.