Question

Let ${\alpha }_1,{\alpha }_2$ and ${\beta }_1,{\beta }_2$ be the roots of $a{x}^2+bx+c=0$ and $p{x}^2+qx+r=0$ 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. $\frac{b^2}{q^2}=\frac{ac}{pr}$
• 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 $a{x}^2+bx+c=0$ and $p{x}^2+qx+r=0$ 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
$\therefore \left|\begin{array}{cc}{\alpha }_1& {\alpha }_2\\ {\beta }_1& {\beta }_2\end{array}\right|=0{\alpha }_1{\beta }_2-{\alpha }_2{\beta }_1=0$
or $\frac{{\alpha }_1}{{\beta }_1}=\frac{{\alpha }_2}{{\beta }_2}=\frac{{\alpha }_1+{\alpha }_2}{{\beta }_1+{\beta }_2}=\sqrt{\frac{{\alpha }_1{\alpha }_2}{{\beta }_1{\beta }_2}}$
$\Rightarrow \frac{pb}{qa}=\sqrt{\frac{pc}{ra}}\Rightarrow \frac{b^2}{q^2}=\frac{ac}{pr}$