Question

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

• A. real and same in sign
• B. real and opposite in sign
• C. equal
• D. data is insufficient

Answer 1

Answer: (B)

real and opposite in sign

Given $\alpha ,\beta$ are the roots of $a{x}^2+bx+c$

So,$\alpha +\beta =\frac{-b}{a},\alpha \beta =\frac{c}{a}$

Similarly, ${\alpha }^4+{\beta }^4=\frac{-m}{l},{\alpha }^4{\beta }^4=\frac{n}{l}$

$f\left(x\right)=a^{2}l{x}^2-4aclx+2c^{2}l+a^{2}m=0$

This can be written as $x^2-4\left(\frac{c}{a}\right)x+(\frac{2c^2}{a^2}+\frac{m}{l})=0$

$=x^2-4\left(\alpha \beta \right)x+(2{\alpha }^2{\beta }^2-{\alpha }^4-{\beta }^4)=x^2-4\alpha \beta x-({\alpha }^2+{\beta }^2{)}^2$

The roots of this equation are,

$\frac{4\alpha \beta \pm \sqrt{16{\alpha }^2{\beta }^2+4({\alpha }^2-{\beta }^2{)}^2}}{2}$

$=\frac{4\alpha \beta \pm 2\sqrt{4{\alpha }^2{\beta }^2+({\alpha }^2-{\beta }^2{)}^2}}{2}=\frac{4\alpha \beta \pm 2\sqrt{({\alpha }^2+{\beta }^2{)}^2}}{2}=\frac{4\alpha \beta \pm 2({\alpha }^2+{\beta }^2)}{2}$

$\alpha \beta =\frac{c}{a}$

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =\frac{b^2}{a^2}-\frac{2c}{a}$

$\frac{4\alpha \beta \pm 2({\alpha }^2+{\beta }^2)}{2}=2\alpha \beta \pm ({\alpha }^2+{\beta }^2)=\frac{2c}{a}\pm (\frac{b^2}{a^2}-\frac{2c}{a})$

roots are $\frac{b^2}{a^2},\frac{4c}{a}-\frac{b^2}{a^2}$

$\frac{b^2}{a^2}>0$

$\frac{4c}{a}-\frac{b^2}{a^2}=\frac{4ac-b^2}{a^2}$

since, $\alpha$,$\beta$ are real and distinct, $4ac-b^2<0$

So, the roots are real and of opposite signs.