Question

kindly anybody please solve this.​

Answer 1

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

Given Polynomial:

$\rm \longrightarrow f(x) = {x}^{2} + px - q$

Let α and β be the zeros of f(x)

Then:

$\rm \longrightarrow \alpha + \beta = - p$

$\rm \longrightarrow \alpha \beta = - q$

Now, it's given that:

$\rm \longrightarrow k = \dfrac{ \alpha }{ \beta }$

Consider the expression given:

$\rm = \dfrac{k}{1 + {k}^{2} }$

$\rm = \dfrac{ \dfrac{ \alpha }{ \beta } }{1 + { \dfrac{ \alpha {}^{2} }{ \beta {}^{2} } }}$

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

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

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

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

$\rm = \dfrac{ - q }{ {p}^{2} + 2q}$

★ 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:

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

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

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:

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

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

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

Answer 2

Answer:

bro is just like your question answer but this is not right answer