Question

If $\alpha ,\beta$ are the roots of $x^{2} +x+1=0$, then $\frac{\alpha }{\beta ^{2} } +\frac{\beta }{\alpha ^{2} } =?$

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given equation:

$\rm:\longmapsto {x}^{2} + x + 1 = 0$

Comparing the equation with ax² + bx + c = 0, we get:

$\rm:\longmapsto \begin{cases} \rm a = 1\\ \rm b = 1 \\ \rm c = 1\end{cases}$

α and β be the roots of this equation.

Therefore:

$\rm: \longmapsto \alpha + \beta = \dfrac{ - b}{a} = - 1$

$\rm: \longmapsto \alpha \beta = \dfrac{c}{a} =1$

Now, we will find the required value.

$\rm = \dfrac{ \alpha }{ { \beta }^{2} } + \dfrac{ \beta }{ { \alpha }^{2} }$

LCM of α² and β² is α²β². Therefore:

$\rm = \dfrac{ { \alpha }^{3} + { \beta }^{3} }{ { \alpha }^{2} { \beta }^{2} }$

As we know that:

$\rm: \longmapsto {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$

$\rm: \longmapsto {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)$

Therefore, the fraction becomes:

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

Now substitute the value in the expression, we get:

$\rm = \dfrac{ {( - 1) }^{3} - 3 \times 1 \times ( - 1) }{1^{2} }$

$\rm = \dfrac{ - 1 + 3}{1}$

$\rm =2$

Therefore:

$\rm: \longmapsto\dfrac{ \alpha }{ { \beta }^{2} } + \dfrac{ \beta }{ { \alpha }^{2} } = 2$

Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

1. Relationship between zeros and coefficients (Quadratic Polynomial).

Let f(x) = ax² + bx + c and let α and β be the zeros of f(x).

Therefore:

$\rm\implies\alpha+\beta=\dfrac{-b}{a}$

$\rm\implies\alpha\beta=\dfrac{c}{a}$

2. Relationship between zeros and coefficients (Cubic Polynomial)

Let f(x) = ax³ + bx² + cx + d and let α, β and γ be the zeros of f(x).

Therefore:

$\rm\implies \alpha+\beta+\gamma=\dfrac{-b}{a}$

$\rm\implies \alpha\beta+\beta\gamma+\alpha\gamma=\dfrac{c}{a}$

$\rm\implies \alpha\beta\gamma=\dfrac{-d}{a}$