How to find out the solution to this problem

Question

Let $ k $ be a real number not equal to 0.

If $ \alpha , \beta $ are non - zero complex numbers, satisfying

$$ \alpha + \beta = -2k $$

$$ \alpha^2 + \beta^2 = 4k^2 - 2k $$

Then I need to find out a quadratic equation having $ \frac{\alpha + \beta}{\alpha} $ and $ \frac{\alpha + \beta}{\beta} $ as it's roots.

How can I find such an quadratic equation? Hints are too welcomed. But I'm a beginner to quadratic equations.

Answer 1

Since $\alpha\beta=\frac{(\alpha+\beta)^2-\alpha^2-\beta^2}{2}=\frac{4k^2-(4k^2-2k)}{2}=k,$ we obtain: $$\frac{\alpha+\beta}{\alpha}+\frac{\alpha+\beta}{\beta}=\frac{\alpha+\beta}{\alpha}\cdot\frac{\alpha+\beta}{\beta}=\frac{4k^2}{k}=4k,$$ which gives the answer: $$z^2-4kz+4k=0.$$

Answer 2

Let $\alpha,\beta$ be the roots of $$t^2+2kt+r=0$$

$\implies\alpha\beta=r$

$y=\dfrac{\alpha+\beta}{\alpha}=\dfrac{-2k}\alpha\iff\alpha=-\dfrac{2k}y$

But as $\alpha$ satisfies $$t^2+2kt+r=0$$

$$\left(-\dfrac{2k}y\right)^2+2k\left(-\dfrac{2k}y\right)+r=0\iff ry^2-4k^2y+4k^2=0$$

Clearly, we shall arrive at the same result if we start with $y=\dfrac{\alpha+\beta}{\beta}$

Now $2r=2\alpha\beta=(\alpha+\beta)^2-(\alpha^2+\beta^2)=(-2k)^2-(4k^2-2k)=2k$