Question

Let $\alpha_1, \alpha_2$ and $\beta_1, \beta_2$ be the roots of $ax^2+ bx+ c = 0$ and $px^2 + qx + r = 0$ respectively. If the system of equations $\alpha_1y + \alpha_2z = 0$ and $\beta_1y + \beta_2z = 0$ has a non-trivial solution, then

• A. $\frac{b ^{2}}{q^{2}} = \frac{ac}{pr} $
• B. $\frac{c^{2}}{r^{2}} = \frac{ab}{pq} $
• C. $\frac{a^{2}}{p^{2}} = \frac{bc}{pr} $
• D. None of these

Answer 1

Answer: (A)

Since $\alpha_1, \alpha_2$ and $ \beta_1, \beta_2$ are the roots of $ax^2 + bx + c = 0$ and $px^2 + qx + r = 0 $ respectively, therefore
$\alpha_{1}+\alpha_{2} = \frac{-b}{a}, \alpha_{1}a_{2} = \frac{c}{a} $ ....(1)
and $\beta_{1} + \beta_{2} = \frac{-q}{p} , \beta_{1}\beta_{2} = \frac{r}{p}$ .....(2)
Since the given system of equation has a non-trivial solution
$ \therefore \begin{vmatrix}\alpha_{1}&\alpha_{2}\\ \beta_{1}&\beta_{2}\end{vmatrix} = 0 \alpha_{1}\beta_{2} - \alpha_{2}\beta_{1} = 0$
or $\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{pb}{qa} = \sqrt{\frac{pc}{ra}} \Rightarrow \frac{b^{2}}{q^{2}} = \frac{ac}{pr} $