Question

If $\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0$ and $\gamma ,\delta$ those of $l{x}^2+mx+n=0$ , then the equation whose roots are $\alpha \gamma +\beta \delta$ and $\alpha \delta +\beta \gamma$ is

• A. $a^2{l}^2{x}^2-ablmx+b^{2}ln+ac{m}^2-4acln=0$
• B. $al{x}^2-ablmx+(a+b-c)(l+m-n)=0$
• C. $a^2{l}^2{x}^2+(a^2+b^2)(l^2+m^2)x-(a+b-c)(l+m-n)=0$
• D. None of these

Answer 1

Answer: (A)

From the given equations
$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a},\gamma +\delta =-\frac{m}{1},\gamma \delta =\frac{n}{1}$
Now, $\left(\alpha \gamma +\beta \delta \right)+\left(\alpha \delta +\beta \gamma \right)=\left(\alpha +\beta \right)\left(\gamma +\delta \right)=\frac{bm}{al}$
and $\left(\alpha \gamma +\beta \delta \right)+\left(\alpha \delta +\beta \gamma \right)={\alpha }^2\gamma \delta +\alpha \beta {\gamma }^2+\alpha \beta {\delta }^2+{\beta }^2\gamma \delta$
$=\left({\alpha }^2+{\beta }^2\right)\gamma \delta +\left({\gamma }^2+{\delta }^2\right)\alpha \beta$
$=\left\{{\left(\alpha +\beta \right)}^2-2\alpha \beta \right\}\gamma \delta +\left\{{\left(\gamma +\delta \right)}^2-2\gamma \delta \right\}\alpha \beta$
$=\left(\frac{b^2}{a^2}-\frac{2c}{a}\right)\frac{n}{1}+\left(\frac{m^2}{1^2}-\frac{2n}{1}\right)\frac{c}{a}=\frac{n\left(b^2-2ac\right)}{a^{2}l}+\frac{c\left(m^2-2nl\right)}{a{l}^2}=0$
$\therefore$ The required equation is,
$x^2-\left\{\left(\alpha \gamma +\beta \delta \right)+\left(\alpha \delta +\beta \gamma \right)\right\}x+\left(\alpha \gamma +\beta \delta \right)+\left(\alpha \delta +\beta \gamma \right)=0$
$\Rightarrow x^2-\frac{bm}{al}x+\frac{ln\left(b^2-2ac\right)+ac\left(m^2-2nl\right)}{a^2{l}^2}=0$
$\Rightarrow a^2{l}^2{x}^2-ablmx+b^{2}ln+ac{m}^2-4acln=0$