Question

If $\alpha$ and $\beta$ be the roots of the equation $a{x}^2+bx+c=0$ find the equation whose roots are
(i) $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$
(ii) $\frac{{\alpha }^2}{\beta }$ and $\frac{{\beta }^2}{\alpha }$

Answer 1

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

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

(i)

$\frac{\alpha }{\beta },\frac{\beta }{\alpha }$

sum of the roots:

$=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

$=\frac{{\alpha }^2+{\beta }^2}{\beta \alpha }$

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

$=\frac{b^2-2ac}{ac}$

product of roots:

$=\frac{\alpha }{\beta }\times \frac{\beta }{\alpha }$

$=1$

$p\left(x\right)=x^2-(sum-of-products)x+product-of-roots$

$=x^2-\left(\frac{b^2-2ac}{ac}\right)x+1$

$\therefore p\left(x\right)=ac{x}^2-(b^2-2ac)x+ac$

(ii)

$\frac{{\alpha }^2}{\beta },\frac{{\beta }^2}{\alpha }$

sum of the roots:

$=\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }$

$=\frac{{\alpha }^3+{\beta }^3}{\beta \alpha }$

$=\frac{(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )}{\alpha \beta }$

$=\frac{-b^3+3abc}{a^{2}c}$

product of roots:

$=\frac{{\alpha }^2}{\beta }\times \frac{{\beta }^2}{\alpha }$

$=\alpha \beta$

$=\frac{c}{a}$

$p\left(x\right)=x^2-(sum-of-products)x+product-of-roots$

$=x^2-\left(\frac{-b^3+3abc}{a^{2}c}\right)x+\frac{c}{a}$

$\therefore p\left(x\right)=a^{2}c{x}^2-(-b^3+3abc)x+a{c}^2$