Question

Let $\alpha _{1},\alpha _{2}$ and $\beta _{1},\beta _{2}$ are roots of the equation $ax^{2}+bx+c=0$ and $px^{2}+qx+r=0$ respectively. If the system of equations $\alpha _{1}y+\alpha _{2}z=0$ and $\beta _{1}y+\beta _{2}z=0$ has a non trivial solution, then

• A. $bpr$ $\_{}^{2}=qac^{2}$
• B. $b^{2}pr=q^{2}ac$
• C. $bp^{2}r=qa^{2}c$
• D. none of these

Answer 1

Answer: (B)

$\begin{vmatrix} \alpha _{1} & \alpha _{2} \\ \beta _{1} & \beta _{2} \end{vmatrix}=0\Rightarrow \alpha _{1}\beta _{2}-\alpha _{2}\beta _{1}=0\Rightarrow \frac{\alpha _{1}}{\alpha _{2}}=\frac{\beta _{1}}{\beta _{2}}$
Using componendo and dividendo $\frac{\alpha _{1} + \alpha _{2}}{\alpha _{1} - \alpha _{2}}=\frac{\beta _{1} + \beta _{2}}{\beta _{1} - \beta _{2}}\Rightarrow \frac{- b / a}{\sqrt{\frac{b^{2}}{a^{2}} - \frac{4 c}{a}}}=\frac{- q / p}{\sqrt{\frac{q^{2}}{p^{2}} - \frac{4 r}{p}}}$
$\Rightarrow \frac{b}{\sqrt{b^{2} - 4 a c}}=\frac{q}{\sqrt{q^{2} - 4 p r}}\Rightarrow b^{2}pr=q^{2}ac$