Question

If $\alpha$ and $\beta$ are the roots of the equation $a{x}^2+bx+c=0$ then the equation whose roots are $({\alpha }^2+{\beta }^2),(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2})$ is

• A. $(acx{)}^2-(b^2-2ac\left)\right(a^2+c^2)x+(b^2-2ac{)}^2=0$
• B. $(acx{)}^2+(b^2-2ac\left)\right(a^2+c^2)x+(b^2-2ac{)}^2=0$
• C. $(acx{)}^2-(b^2-2ac\left)\right(a^2+c^2)x-(b^2-2ac{)}^2=0$
• D. $(acx{)}^2+(b^2-2ac\left)\right(a^2+c^2)x-(b^2-2ac{)}^2=0$

Answer 1

The correct option is A $(acx{)}^2-(b^2-2ac\left)\right(a^2+c^2)x+(b^2-2ac{)}^2=0$

$a{x}^2+bx+c=0$
Sum and product of roots is,
$\alpha +\beta =-\frac{b}{a}$ & $\alpha \beta =\frac{c}{a}$
Let $S$ be the sum and $P$ be the product of the roots of required equation,
Now,
$S=({\alpha }^2+{\beta }^2)+\frac{{\alpha }^2+{\beta }^2}{(\alpha \beta {)}^2}=(\alpha +\beta {)}^2-2\alpha \beta +\frac{(\alpha +\beta {)}^2-2\alpha \beta }{(\alpha \beta {)}^2}=\left(\frac{b^2-2ac}{a^2}\right)+\left(\frac{b^2-2ac}{c^2}\right)=\frac{(b^2-2ac)(a^2+c^2)}{a^2{c}^2}$
And
$P=\frac{({\alpha }^2+{\beta }^2{)}^2}{{\alpha }^2{\beta }^2}={\left(\frac{b^2-2ac}{ac}\right)}^2$

Hence, the required equation is
$x^2-Sx+P=0$
$\Rightarrow (acx{)}^2-(b^2-2ac\left)\right(a^2+c^2)x+(b^2-2ac{)}^2=0$