My question is derivative of this one here that caused me to get an entirely new question that came out of my working on it and reading some very helpful answers.
Two quadratic equations have real roots $\alpha$ and $\beta$ such that $$\alpha - \beta = 3$$ and $$\alpha \beta = 2(\alpha + \beta).$$ Find the two possible quadratic equations that satisfy these conditions.
Finding $\alpha$ or $\beta$ can be found since we have 2 equations and 2 unknowns.
\begin{align}\alpha &= \beta + 3\\ (\beta +3)(\beta) &= 2(\beta + 3 + \beta)\\ \beta ^2 + 3\beta &= 4\beta + 6\\ \beta ^2 - \beta - 6 &= 0\\ \beta_1 &= 3\\ \beta_2 &= -2\end{align}
And thus, using our relations:
\begin{align}\alpha_1 &= 6\\ \alpha_2 &= 1\end{align}
And thus, we have the two sets of roots for the two equations. My question, however, is not about the maths in this question but about the maths itself, and excuse me for my possible vagueness in the following question - but how does what I've done manage to find solutions to both sets of $\alpha$ and $\beta$ values that satisfies the two equations?
How is it that, since I've created a quadratic, I've been able to find the two values of $\beta$ that are the specific roots to the two quadratic equations I want rather than having a potential valid $\beta$ value for one, and an invalid one for the other? How is it that both were valid and not just one? Why does this quadratic account for both equations? For example, if I had a third equation relating $\alpha$ and $\beta$, could I find a third quadratic? The question is simply why the two values for $\beta$ and therefore $\alpha$ were both the values for the two quadratics rather than, say, one of the solutions not being valid?