Question

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

• A. $a^2l^2 x^2 - ablmx + b^2l n+ acm^2 - 4acl\, n = 0$
• B. $alx^2 -ablmx + (a +b -c)(l +m-n) = 0$
• C. $a^2l^2x^2 + (a^2 +b^2 )(l^2 +m^2 ) x -(a +b - c)(l +m-n) = 0$
• 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 $\left(\alpha \gamma +\beta \delta \right) + \left(\alpha \delta +\beta \gamma \right) = \alpha^{2}\gamma\delta + \alpha\beta\gamma^{2} + \alpha\beta\delta^{2} + \beta^{2}\gamma\delta$
$= \left(\alpha^{2} + \beta^{2} \right)\gamma\delta+\left(\gamma^{2}+\delta^{2}\right)\alpha\beta$
$= \left\{\left(\alpha+\beta\right)^{2}-2\alpha\beta\right\}\gamma\delta+\left\{\left(\gamma+\delta\right)^{2}-2\gamma\delta\right\}\alpha\beta$
$= \left(\frac{b^{2}}{a^{2}}-\frac{2c}{a}\right) \frac{n}{1} + \left(\frac{m^{2}}{1^{2}}-\frac{2n}{1}\right) \frac{c}{a} = \frac{n\left(b^{2}-2ac\right)}{a^{2}l} + \frac{c\left(m^{2}-2nl\right)}{al^{2}} = 0$
$\therefore \quad$ The required equation is,
$x^{2} - \left\{\left(\alpha \gamma +\beta \delta \right) + \left(\alpha \delta +\beta \gamma \right)\right\}x+\left(\alpha \gamma +\beta \delta \right) + \left(\alpha \delta +\beta \gamma \right) = 0$
$\Rightarrow x^{2}- \frac{bm}{al}x +\frac{ln\left(b^{2}-2ac\right)+ac\left(m^{2}-2nl\right)}{a^{2}l^{2}}=0$
$\Rightarrow a^{2} l^{2} x^{2} - ablmx + b^{2} ln+acm^{2} - 4acln = 0$