Question

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

• A. real and opposite in sign
• B. equal
• C. imaginary
• D. None of these

Answer 1

The correct option is A real and opposite in sign

We have,
$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$
and ${\alpha }^4+b^4=-\frac{m}{l},{\alpha }^4{\beta }^4=\frac{n}{l}$
The given equation $a^{2}l{x}^2-4aclx+2c^{2}l+a^{2}m=0$
has discriminant $D={16a}^2{c}^2{l}^2-4a^{2}l(2c^{2}l+a^{2}m)=8a^2{c}^2{l}^2-4a^{4}lm$
$=4a^4{l}^2(\frac{2c^2}{a^2}-\frac{m}{l})>0$ $[\frac{-m}{l}={\alpha }^4+{\beta }^4>0]$
Hence, the roots are real.
Product of the roots $=\frac{2c^{2}l+a^{2}m}{a^{2}l}$
$=\frac{2c^2}{a^2}+\frac{m}{l}=2{\alpha }^2{\beta }^2-({\alpha }^4+{\beta }^4)=-{({\alpha }^2+{\beta }^2)}^2<0$
$\therefore$The roots are of opposite signs.