Mathematics - Determinant Question with Solution | TestHub

Since ${\alpha }_1,{\alpha }_2$ and ${\beta }_1,{\beta }_2$ are the roots of ax ${}^2$ $+bx+c=0$ and $p{x}^2+qx+r=0$ 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 $\frac{{\alpha }_1}{{\beta }_1}=\frac{{\alpha }_2}{{\beta }_2}=\frac{{\alpha }_1+{\alpha }_2}{{\beta }_1+{\beta }_2}=\sqrt{\frac{{\alpha }_1{\alpha }_2}{{\beta }_1{\beta }_2}}$ $\Rightarrow \frac{\mathrm{pb}}{\mathrm{qa}}=\sqrt{\frac{\mathrm{pc}}{\mathrm{ra}}}\Rightarrow \frac{{\mathrm{b}}^2}{{\mathrm{q}}^2}=\frac{\mathrm{ac}}{\mathrm{pr}}$