Mathematics - Quadratic Equation Question with Solution | TestHub

Sum of roots $=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$ and product $=1$ Given, $\alpha +\beta =-p$ and ${\alpha }^3+{\beta }^3=q$ $$\begin{array}{rl}& \Rightarrow (\alpha +\beta )\left({\alpha }^2-\alpha \beta +{\beta }^2\right)=q\\ & \therefore {\alpha }^2+{\beta }^2-\alpha \beta =\frac{-q}{p}\end{array}$$ and $(\alpha +\beta {)}^2=p^2$ $$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =p^2$$ From Eqs. (i) and (ii), we get $${\alpha }^2+{\beta }^2=\frac{p^3-2q}{3p}$$ and $\alpha \beta =\frac{p^3+q}{3p}$ $\therefore$ Required equation is $$\begin{array}{c}{x}^2-\frac{\left(p^3-2q\right)x}{\left(p^3+q\right)}+1=0\\ \Rightarrow \left(p^3+q\right)x^2-\left(p^3-2q\right)x+\left(p^3+q\right)=0\end{array}$$