Question

Let ${\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2$ be the roots of $a{x}^2+bx+c=0andp{x}^2+qx+r=0$, respectively. If the system of equations ${\alpha }_{1}y+{\alpha }_{2}z=0and{\beta }_{1}y+{\beta }_{2}z=0$ has a non - trivial solution, then

• A. $\frac{b^2}{q^2}=\frac{ac}{pr}$
• B. $\frac{b^2}{q^2}=\frac{pr}{ac}$
• C. $\frac{b}{q}=\frac{ac}{pr}$
• D. $\frac{b^3}{q^3}=\frac{ac}{pr}$

Answer 1

The correct option is A

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

Since, ${\alpha }_1,{\alpha }_2$ are the roots pf $a{x}^2+bx+c=0$.
$\Rightarrow {\alpha }_1+{\alpha }_2=-\frac{b}{a}and{\alpha }_1{\alpha }_2=\frac{c}{a}$ . . . (i)
Also, ${\beta }_1,{\beta }_2$ are the roots of $p{x}^2+qx+r=0$
$\Rightarrow {\beta }_1+{\beta }_2=-\frac{q}{p}and{\beta }_1{\beta }_2=\frac{r}{p}$ . . . (ii)
Given system of equations
${\alpha }_{1}y+{\alpha }_{2}z=0$
and ${\beta }_{1}y+{\beta }_{2}z=0$, has non - trivial solution.
$\therefore \left|\begin{array}{cc}{\alpha }_1& {\alpha }_2\\ {\beta }_1& {\beta }_2\end{array}\right|=0\Rightarrow \frac{{\alpha }_1}{{\alpha }_2}=\frac{{\beta }_1}{{\beta }_2}$
Applying componendo - dividendo
$\frac{{\alpha }_1+{\alpha }_2}{{\alpha }_1-{\alpha }_2}=\frac{{\beta }_1+{\beta }_2}{{\beta }_1-{\beta }_2}\Rightarrow ({\alpha }_1+{\alpha }_2)({\beta }_1-{\beta }_2)=({\alpha }_1-{\alpha }_2)({\beta }_1+{\beta }_2)\Rightarrow ({\alpha }_1+{\alpha }_2{)}^2\{({\beta }_1+{\beta }_2{)}^2-4{\beta }_2{\beta }_2\}=({\beta }_1+{\beta }_2{)}^2\{({\alpha }_1+{\alpha }_2{)}^2-4{\alpha }_1{\alpha }_2\}$
From Eqs. (i) and (ii), we get
$\frac{b^2}{a^2}(\frac{q^2}{p^2}-\frac{4r}{p})=\frac{q^2}{p^2}(\frac{b^2}{a^2}-\frac{4c}{a})\Rightarrow \frac{b^{2}r}{a^{2}p}=\frac{q^{2}c}{a{p}^2}\Rightarrow \frac{b^{2}r}{a}=\frac{q^{2}c}{p}\Rightarrow \frac{b^2}{q^2}=\frac{ac}{pr}$