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. $b^{2}pr=q^{2}ac$
• B. $bp{r}^2=qa{c}^2$
• C. $b{p}^{2}r=q{a}^{2}c$
• D. None of these

Answer 1

The correct option is A $b^{2}pr=q^{2}ac$

Given system has non-trivial solution, then

$\left|\begin{array}{cc}{\alpha }_1& {\alpha }_2\\ {\beta }_1& {\beta }_2\end{array}\right|=0$

$\Rightarrow {\alpha }_1{\beta }_2={\alpha }_2{\beta }_1$

$\Rightarrow \frac{{\alpha }_1}{{\beta }_1}=\frac{{\alpha }_2}{{\beta }_2}=\frac{{\alpha }_1+{\alpha }_2}{{\beta }_1+{\beta }_2}=\sqrt{\frac{{\alpha }_1{\alpha }_2}{{\beta }_1{\beta }_2}}$

$\Rightarrow \frac{-b/a}{-q/p}=\sqrt{\frac{c/a}{r/p}}$

$\Rightarrow \frac{bp}{aq}=\sqrt{\frac{cp}{ar}}$

$\Rightarrow b^2{p}^{2}ar=a^2{q}^{2}cp$

$\Rightarrow b^{2}pr=q^{2}ac$