Mathematics - Quadratic Equation Question with Solution | TestHub

For the equation $x^2-6x+3=0$, by Vieta's formulas, the sum of the roots $\alpha +\beta =-(-6)/1=6$ and the product of the roots $\alpha \beta =3/1=3$. We need to find the value of $\frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}$. Combining the terms, we get $\frac{{\alpha }^3+{\beta }^3}{(\alpha \beta {)}^2}$. We know that ${\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$. Substituting the values, ${\alpha }^3+{\beta }^3=(6{)}^3-3(3)(6)=216-54=162$. Therefore, $\frac{{\alpha }^3+{\beta }^3}{(\alpha \beta {)}^2}=\frac{162}{(3{)}^2}=\frac{162}{9}=18$.