# How is a quadratic equation transformed when one or both of its roots tend to infinity?

We have a quadratic equation $$y=ax^2+bx+c$$

With roots $$\alpha,\beta$$

Case 1

$$\alpha\to\infty$$ and $$\beta$$ is a finite number the equation transforms to $$y_1=bx+c$$
Case 2

$$\alpha,\beta\to\infty$$ the new equation becomes
$$y_2=c$$

Case 3

$$\alpha\to\infty,\beta\to -\infty$$

$$y_3=??$$

I figured out how the equation transforms in the first two cases by tinkering around on a graphing calculator but when I had to prove it rigorously using math I treated as a problem in limits but it looks like a problem in multivariable calculus? Since $$a,b,c$$ are dependent on each other and I haven't learnt that yet.

Is there some other way to figure out these transformations without using math above High School level?

## 1 Answer

By factoring, your equation is immediately converted to $$y = a(x - \alpha)(x - \beta)$$ which can then be re-expanded to give $$y = ax^2 - a(\alpha+\beta)x + a \alpha\beta$$

As you can see, the constant $$a$$ is not determined by $$\alpha$$ and $$\beta$$, and so this problem is underdetermined. One way to deal with that is to choose $$a$$ in some appropriate manner, depending on the case we are in.

For instance, in Case 1 we could choose $$a = \frac{1}{\alpha}$$ and then the equation becomes $$y = \frac{1}{\alpha} x^2 - (1 + \frac{\beta}{\alpha}) x + \beta$$ Now, taking a limit as $$\alpha \to +\infty$$ the equation becomes $$y = -x + \beta$$ whose only root is $$\beta$$. Good so far.

In Case 2, we could choose $$a = \frac{1}{\alpha\beta}$$ and the equation becomes $$y = \frac{1}{\alpha\beta}x^2 - \left(\frac{1}{\beta} + \frac{1}{\alpha}\right) x + 1$$ Again, taking a limit as $$\alpha,\beta \to +\infty$$, the equation becomes $$y = 1$$

We can deal with Case 3 in the same fashion as Case 2, with the exact same outcome: $$y = 1$$

• Why can we assign values to $a$ which are dependent on $\alpha,\beta$ wouldn't doing so mean that it's valid only for a specific case? – user659291 Apr 15 at 2:50
• You've essentially done the same thing, but you called it "transforming the equation". So you could similarly ask yourself: Why can you transform the equation? – Lee Mosher Apr 15 at 10:38