Question

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

• A. $a{c}^2{x}^2+(3abc-b^3)x+a^{2}c=0$
• B. $a{c}^2{x}^2+(3abc-b^3)x-a^{2}c=0$
• C. $a{c}^2{x}^2-(3abc-b^3)x+a^{2}c=0$
• D. $a{c}^2{x}^2-(3abc-b^3)x-a^{2}c=0$

Answer 1

Answer: (C)

$a{c}^2{x}^2-(3abc-b^3)x+a^{2}c=0$

$\alpha +\beta =$ sum of roots $=\frac{-b}{a}$
$\alpha \beta =\frac{c}{a.}$

Let $\frac{\alpha }{{\beta }^2},\frac{\beta }{{\alpha }^2},$ be roots
$\Rightarrow$ sum of roots $=\frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}=\frac{{\alpha }^3+{\beta }^3}{{\alpha }^2{\beta }^2}$

$=\frac{(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )}{{\alpha }^2{\beta }^2}$

$=\frac{(\frac{-b}{a}{)}^3-3\frac{c}{a}(\frac{-b}{a})}{(\frac{c}{a}{)}^2}$

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

Product of roots $=\frac{\alpha }{{\beta }^2}\times \frac{\beta }{{\alpha }^2}=\frac{1}{\alpha \beta }=\frac{a}{c}$

$\therefore$ equation is $x^2-\frac{(3abc-b^3)x}{a{c}^2}+\frac{a}{c}=0$

$\Rightarrow a{c}^2{x}^2-[3abc-b^3]x+a^{2}c=0.$