Question

Let $\alpha ,\beta$ be the roots of the equation $a{x}^2+bx+c=0$ and ${\alpha }^4+{\beta }^4$ be the roots of the equation $p{x}^2+qx+r=0$, then the roots of the equation $a^{2}p{x}^2-4acpx+2c^{2}p+a^{2}q=0$ are:

• A. Always +ve
• B. Always complex
• C. Opposite in sign
• D. Negative

Answer 1

The correct option is C Opposite in sign

$a{x}^2+bx+c=0$ ......... $\left(i\right)$
$\alpha ,\beta$ are the roots of above equation
$\therefore \alpha +\beta =\frac{-b}{a}$ ..... $\left(ii\right)$
$\alpha \beta =\frac{c}{a}$ ........ $\left(iii\right)$

${\alpha }^4$ and ${\beta }^4$ are the roots of equation $p{x}^2+qx+r=0$
$\therefore {\alpha }^4+{\beta }^4=\frac{-q}{p}$ .... $\left(iv\right)$
${\alpha }^4{\beta }^4=\frac{r}{p}$ ....... $\left(v\right)$
$a^{2}p{x}^2-4acpx+2c^{2}p+a^{2}q=0$
Let $r$ and $\delta$ be roots of above equation
$\therefore r+\delta =\frac{4acp}{a^{2}p}=\frac{4c}{a}$
$r\delta =\frac{2c^{2}p+a^{2}q}{a^{2}p}$
$=2{\left(\frac{c}{a}\right)}^2+\frac{q}{p}$

$=2{\alpha }^2{\beta }^2-({\alpha }^4+{\beta }^4)$ ..... [From equation $\left(iii\right)$ and $\left(iv\right)$]
$=-({\alpha }^4+{\beta }^4-2{\alpha }^2{\beta }^2)$
$=-{({\alpha }^2-{\beta }^2)}^2$
${({\alpha }^2-{\beta }^2)}^2>0$
$\therefore -{({\alpha }^2-{\beta }^2)}^2<0$
$\therefore r\delta <0$

$\Rightarrow (r<0\&\delta >0)$ or $(r>0\&\delta <0)$
$\therefore$ Roots of given equation are opposite in sign.