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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?

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I think your questions come from the formulation of the problem, which might be a bit confusing. There are actually two problems in one:

  1. Make a quadratic equation with roots $a_i$ and $b_i$.

  2. Find the solutions $(\alpha_i,\beta_i)$ that satisfy the two equations $$ \alpha - \beta = 3 \\ \alpha \beta = 2 (\alpha + \beta) $$

The problem asks you to find solutions $(\alpha_i,\beta_i)$ and then use those solutions to form quadratic equations in step 1.

The first bit is easy, because a quadratic equation with two roots $a_i$ and $b_i$ is always of the form: $$ f_i(x) = c_i (x - a_i) (x - b_i) $$ where the constant $c_i$ can be anything non-zero.

So we are left with finding numbers $(\alpha,\beta)$ that satisfy the two given equations. How many such pairs can be found is not immediately obvious and depends on the type of equations you have. In this case the problem already informed you that you will need to find two.

Luckily in this case the first one is linear and you immediately find that in any case $\alpha=\beta+3$. And substitution in the second gives a single quadratic equation in $\beta$ that you can solve. You also know that there would be at most two possible distinct answers for $\beta$ and each has a corresponding $\alpha$.

Could you have find out that there are two solutions without the calculation? Yes, you could have made a graph where you put $\alpha$ on the $x$-axis, and $\beta$ on the $y$-axis. Then the first equation gives you a function: $$ \beta(\alpha) = \alpha - 3 $$ which is a straight line in the graph and shows you all points $(\alpha,\beta)$ that satisfy the first equation. The second equation would give you $$ \beta(\alpha) = \frac{2 \alpha}{\alpha-2} $$ and if you would plot this in the graph you have two curves, one for $\alpha<2$ and one for $\alpha>2$ and they give you all points $(\alpha,\beta)$ that satisfy the second equation.

In order for both equation to be satisfied at the same time, we are therefore looking at when these different curves cross and this occurs in only two points.

If you would have added an additional equation to the problem and hence a third curve in the graph it would in general not go to either of those points and hence it would not be possible to satisfy the three equations at the same time.

It is important to realise that the graphing approach I sketched above only works under the assumption that $\alpha$ and $\beta$ are real numbers and that there might be additional solutions in complex numbers.

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