[tex]\large\underline{\sf{Solution-}}[/tex]
Given Polynomial:
[tex] \rm \longmapsto f(x) = {x}^{2} + px - q[/tex]
Let α and β be the zeros of f(x)
Then:
[tex] \rm \longmapsto \alpha + \beta = - p[/tex]
[tex] \rm \longmapsto \alpha \beta = - q[/tex]
Now, it's given that:
[tex] \rm \longmapsto k = \dfrac{ \alpha }{ \beta } [/tex]
Consider the expression given:
[tex] \rm = \dfrac{k}{1 + {k}^{2} } [/tex]
[tex] \rm = \dfrac{ \dfrac{ \alpha }{ \beta } }{1 + { \dfrac{ \alpha {}^{2} }{ \beta {}^{2} } }} [/tex]
[tex] \rm = \dfrac{ \dfrac{ \alpha }{ \beta } }{ { \dfrac{ \alpha {}^{2} + { \beta }^{2} }{ \beta {}^{2} } }} [/tex]
[tex] \rm = \dfrac{ \alpha }{ { \dfrac{ \alpha {}^{2} + { \beta }^{2} }{ \beta } }} [/tex]
[tex] \rm = \dfrac{ \alpha \beta }{\alpha {}^{2} + { \beta }^{2} } [/tex]
[tex] \rm = \dfrac{ \alpha \beta }{ {( \alpha + \beta )}^{2} - 2 \alpha \beta } [/tex]
[tex] \rm = \dfrac{ - q }{ {p}^{2} + 2q} [/tex]
