Question

Find a quadratic equation whose roots are $\alpha$ and $\beta$ such that $\alpha +\beta =3$ and ${\alpha }^3+{\beta }^3=9$.

Answer 1

Step-1: Solve for the value of $\alpha \beta$ from the given expressions:

Using identity ${\left(\alpha +\beta \right)}^{\mathbf{3}}\mathbf{=}{\mathit{\alpha }}^{\mathbf{3}}\mathbf{+}{\mathit{\beta }}^{\mathbf{3}}\mathbf{+}\mathbf{3}\mathit{\alpha }\mathit{\beta }\left(\alpha +\beta \right)$.

$$\begin{gathered} \Rightarrow {\left(3\right)}^3=9+3\alpha \beta \left(3\right) \\ \Rightarrow 27=9+9\alpha \beta \\ \Rightarrow 9\alpha \beta =18 \\ \Rightarrow \alpha \beta =2 \end{gathered}$$

Step 2: Put this value in the standard form of quadratic equation:

Quadratic Equation whose roots are $\alpha$and $\beta$is ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\left(\alpha +\beta \right)\mathit{x}\mathbf{+}\mathit{\alpha }\mathit{\beta }\mathbf{=}\mathbf{0}$

$\Rightarrow x^2-3x+2=0$

Hence, the required quadratic equation is ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{0}$