Question

If $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$, form the equation whose roots are ${\alpha }^2+{\beta }^2$ and ${\alpha }^{-2}+{\beta }^{-2}$.

Answer 1

Given that $\alpha ,\beta$ are roots of equation $a{x}^2+bx+c=0$

We have $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$

Noe let the equation whose roots are ${\alpha }^2+{\beta }^2$ and $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$ be $x^2+px+q=0$

Sum of roots is $-p={\alpha }^2+{\beta }^2+\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=({\alpha }^2+{\beta }^2)(1+\frac{1}{(\alpha \beta {)}^2})=\left(\frac{b^2-2ac}{a^2}\right)(1+\frac{a^2}{c^2})=\frac{{(b}^2-2ac\left)\right(c^2+a^2)}{{\left(ac\right)}^2}$

$\Rightarrow p=\frac{(2ac-b^2)(c^2+a^2)}{{\left(ac\right)}^2}$

Product of roots is $q=({\alpha }^2+{\beta }^2)(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2})=\frac{{\left(\frac{b^2-2ac}{a^2}\right)}^2}{\frac{c^2}{a^2}}=\frac{{(b^2-2ac)}^2}{a^2{c}^2}$

Therefore the required equation is $a^2{c}^2{x}^2+(2ac-b^2)(c^2+a^2)x+{(b^2-2ac)}^2=0$