Question

if ${\alpha }_1,{\alpha }_2$ are the roots of the equation $x^2+px+q=0$ and ${\beta }_1,{\beta }_2$ are those of the equation $x^2+rx+s=0$ and the system of the equation ${\alpha }_{1}y+{\alpha }_{2}z=0$ and ${\beta }_{1}y+{\beta }_{2}z=0$ has non trivial solution, then

• A. $\frac{q^2}{r^2}=\frac{p}{s}$
• B. $\frac{p^2}{q^2}=\frac{r}{s}$
• C. $\frac{p^2}{s^2}=\frac{q}{r}$
• D. $\frac{p^2}{r^2}=\frac{q}{s}$

Answer 1

The correct option is D $\frac{p^2}{r^2}=\frac{q}{s}$

Given ${\alpha }_1,{\alpha }_2$ are the roots of the equation $x^2+px+q=0$ and ${\beta }_1,{\beta }_2$ are roots of the equation $x^2+rx+s=0$ and the system of the equation ${\alpha }_{1}y+{\alpha }_{2}z=0$ and ${\beta }_{1}y+{\beta }_{2}z=0$ has non trivial solution.
$\Delta =0$
$\Rightarrow \left|\begin{array}{cc}{\alpha }_1& {\alpha }_2\\ {\beta }_1& {\beta }_2\end{array}\right|=0$
$\Rightarrow {\alpha }_1{\beta }_2-{\alpha }_2{\beta }_1=0$
$\Rightarrow \frac{{\alpha }_1}{{\alpha }_2}=\frac{{\beta }_1}{{\beta }_2}$
Applying componendo and dividendo rule
$\Rightarrow \frac{{\alpha }_1+{\alpha }_2}{{\alpha }_1-{\alpha }_2}=\frac{{\beta }_1+{\beta }_2}{{\beta }_1-{\beta }_2}$
$\Rightarrow \frac{-p}{\sqrt{p^2-4q}}=\frac{-r}{\sqrt{r^2-4s}}$
Squaring on both sides and simplifying gives
$\frac{p^2}{r^2}=\frac{q}{s}$
Hence, option D.