Mathematics - Quadratic Equation Question with Solution | TestHub

Sum of the roots:

$\alpha +{\alpha }^2=-\frac{p}{3}$

Product of the roots:

${\alpha }^3=1$

The solutions for ${\alpha }^3=1$ are the cube roots of unity.

These are $\alpha =1$, $\alpha =e^{i\frac{2\pi }{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, and $\alpha =e^{i\frac{4\pi }{3}}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$.

Case 1: If $\alpha =1$

$p=-6$

However, the problem states that $p>0$. So, this case is not valid.

Case 2: If $\alpha =-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ (which is $\omega$, a complex cube root of unity

$p=3$

This value of $p=3$ satisfies the condition $p>0$.

Case 3: If $\alpha =-\frac{1}{2}-i\frac{\sqrt{3}}{2}$ (which is ${\omega }^2$)

$p=3$

This value of $p=3$ also satisfies the condition $p>0$.