Question

If $\alpha ,\beta$ be two zeroes of the quadratic polynomial $a{x}^2+bx-c=0$, then $\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=$ _______.

• A. $\frac{b^2-2ac}{a^2}$
• B. $\frac{3abc-b^3}{c^3}$
• C. $\frac{3abc-b^3}{a^{2}c}$
• D. $\frac{b^3+3abc}{a^{2}c}$

Answer 1

The correct option is D $\frac{b^3+3abc}{a^{2}c}$

Sum of the roots of the given equation $=\alpha +\beta =\frac{-b}{a}$ ...(1)

Product of the roots of the given equation $=\alpha \beta =-\frac{c}{a}$ ....(2)

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

Substituting (1),(2) in (3), we get

$\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=\frac{(-b/a)\left(\right(-b/a{)}^2-3(-c/a))}{-c/a}=\frac{b}{c}[\frac{b^2}{a^2}+\frac{3c}{a}]=\frac{b}{c}\frac{b^2+3abc}{a^2}=\frac{b^3+3abc}{a^{2}c}$