Q56 / 125 · BITSAT 2016 · Mathematics
Let $\alpha_{1}, \alpha_{2}$ and $\beta_{1}, \beta_{2}$ be the roots of $a x^{2}+b x+c$ $=0$ and $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0$ respectively. If the system of equations $\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}=0$ and $\beta_{1} \mathrm{y}+\beta_{2} \mathrm{z}=0$ has a non-trivial solution, then
- Correct answerA$\frac{b^{2}}{q^{2}}=\frac{a c}{p r}$
- B$\frac{\mathrm{c}^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{ab}}{\mathrm{pq}}$
- C$\frac{a^{2}}{p^{2}}=\frac{b c}{q r}$
- DNone of these
Solution
Since $\alpha_{1}, \alpha_{2}$ and $\beta_{1}, \beta_{2}$ are the roots of ax $^{2}$ $+b x+c=0$ and $p x^{2}+q x+r=0$ respectively
therefore
$$
\alpha_{1}+\alpha_{2}=\frac{-b}{a}, \alpha_{1} \alpha_{2}=\frac{c}{a}
$$
and $\beta_{1}+\beta_{2}=\frac{-q}{p}, \beta_{1} \beta_{2}=\frac{r}{p}$
Since the given system of equation has a non-trivial solution
$$
\therefore\left|\begin{array}{ll}
\alpha_{1} & \alpha_{2} \\
\beta_{1} & \beta_{2}
\end{array}\right|=0 \text { i.e. } \alpha_{1} \beta_{2}-\alpha_{2} \beta_{1}=0
$$
or $\quad \frac{\alpha_{1}}{\beta_{1}}=\frac{\alpha_{2}}{\beta_{2}}=\frac{\alpha_{1}+\alpha_{2}}{\beta_{1}+\beta_{2}}=\sqrt{\frac{\alpha_{1} \alpha_{2}}{\beta_{1} \beta_{2}}}$
$\Rightarrow \frac{\mathrm{pb}}{\mathrm{qa}}=\sqrt{\frac{\mathrm{pc}}{\mathrm{ra}}} \Rightarrow \frac{\mathrm{b}^{2}}{\mathrm{q}^{2}}=\frac{\mathrm{ac}}{\mathrm{pr}}$
Hence, the correct answer is option: 'A'.