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 which of the following option is CORRECT ?

• A. $abc=pqr$
• B. $a^{2}pr=q^{2}bc$
• C. $b^{2}pr=q^{2}ac$
• D. $a^{2}qr=p^{2}bc$

Answer 1

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

$a{x}^2+bx+c=0$
$\Rightarrow {\alpha }_1+{\alpha }_2=\frac{-b}{a},{\alpha }_1{\alpha }_2=\frac{c}{a}$
and $p{x}^2+qx+r=0$
$\Rightarrow {\beta }_1+{\beta }_2=-\frac{q}{p},{\beta }_1{\beta }_2=\frac{r}{p}$

${\alpha }_{1}y+{\alpha }_{2}z=0,{\beta }_{1}y+{\beta }_{2}z=0$ have a non-trivial solution.
$\therefore \left|\begin{array}{cc}{\alpha }_1& {\alpha }_2\\ {\beta }_1& {\beta }_2\end{array}\right|=0$

$\Rightarrow {\alpha }_1{\beta }_2-{\alpha }_2{\beta }_1=0$
$\Rightarrow \frac{{\alpha }_1}{{\beta }_1}=\frac{{\alpha }_2}{{\beta }_2}=k\left(\text{say}\right)$

Now, ${\alpha }_1+{\alpha }_2=\frac{-b}{a}$
$\Rightarrow k({\beta }_1+{\beta }_2)=\frac{-b}{a}$
$\Rightarrow k\left(\frac{-q}{p}\right)=\frac{-b}{a}$
$\Rightarrow k=\frac{pb}{qa}\cdots \left(1\right)$

${\alpha }_1{\alpha }_2=\frac{c}{a}$
$\Rightarrow k^2{\beta }_1{\beta }_2=\frac{c}{a}$
$\Rightarrow k^2\left(\frac{r}{p}\right)=\frac{c}{a}$
$\Rightarrow k^2=\frac{pc}{ar}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$\frac{p^2{b}^2}{q^2{a}^2}=\frac{pc}{ar}$
$\Rightarrow b^{2}pr=q^{2}ac$