Question

Let $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0,(b\ne 0)$ and ${\alpha }^4,{\beta }^4$ are the roots of the equation $l{x}^2+mx+n=0$. If $\alpha ,\beta$ are real and distinct, then the roots of the equation $a^{2}l{x}^2-4aclx+2c^{2}l+a^{2}m=0$ are

• A. always real
• B. opposite in sign
• C. same in sign
• D. always imaginary

Answer 1

Answer: (A), (B)

opposite in sign

$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$
${\alpha }^4+{\beta }^4=-\frac{m}{l},{\alpha }^4{\beta }^4=\frac{n}{l}$
Now,
${\alpha }^4+{\beta }^4=-\frac{m}{l}$
$\Rightarrow ({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2=-\frac{m}{l}$
$\Rightarrow \left(\right(\alpha +\beta {)}^2-2\alpha \beta {)}^2-2{\alpha }^2{\beta }^2=-\frac{m}{l}$
$\Rightarrow {(\frac{b^2}{a^2}-\frac{2c}{a})}^2-\frac{2c^2}{a^2}=-\frac{m}{l}$
$\Rightarrow {\left(\frac{b^2}{a^2}\right)}^2-4\left(\frac{b^2}{a^2}\right)\left(\frac{c}{a}\right)+2\left(\frac{c^2}{a^2}\right)+\frac{m}{l}=0$
$\Rightarrow a^2{\left(\frac{b^2}{a^2}\right)}^2-4ac\left(\frac{b^2}{a^2}\right)+2c^2+\frac{a^{2}m}{l}=0$
$\Rightarrow a^{2}l{\left(\frac{b^2}{a^2}\right)}^2-4acl\left(\frac{b^2}{a^2}\right)+2c^{2}l+a^{2}m=0$
Therefore,
$\frac{b^2}{a^2}$ is a root of the equation
$a^{2}l{x}^2-4aclx+2c^{2}l+a^{2}m=0$

Also, $\frac{b^2}{a^2}>0$
Thus, the equation $a^{2}l{x}^2-4aclx+2c^{2}l+a^{2}m=0$ has a positive real root.
Let $\gamma$ be the other root of this equation.
Then, $\gamma +\frac{b^2}{a^2}=\frac{4acl}{a^{2}l}$
$\Rightarrow \gamma =\frac{4c}{a}-\frac{b^2}{a^2}=\frac{4ac-b^2}{a^2}$
$\Rightarrow \gamma <0$
Hence, roots are always real and opposite in sign.