Roots of quadratic equation $ax^2+bx+c$ at infinity

Question

Our maths teacher taught us that conditions for roots of a quadratic equation $ax^2+bx+c=0$ to lie at infinity are :

A) (for exactly one root at infinity) $a=0$, $b$ = non-zero, & $c$ can be zero or non-zero

B) (for both roots at infinity) $a = 0$, $b= 0$, and $c$ = non-zero

By equal to zero, I mean the coefficients are so small that they are nearly zero, i.e approaching zero. They can’t be exactly equal to zero right? Or it would no longer be a quadratic. Please correct me if I’m wrong.

This is not my question though. What I want to ask is, are these the only two conditions for roots of a quadratic equation to lie at infinity?

For example, if we consider the equations

1) $x^2-4\times 10^{30} = 0$

2) $0.00000001x^2+ 0.0000005x + 6 = 0$

I have attached two screenshots. In the first equation, both roots are at infinity even though coefficient of $x^2$ is non-zero (contrary to what condition (A) says.

In the second equation, none of the roots are lying at infinity even thought coefficients of $x^2$ and $x$ are nearly equal to zero, which again is not consistent with what condition (B) says.

How do I explain these discrepancies?

(Looks like I can’t attach more than one pic here. So I’m attaching just one, graph of equation (2). It’s clear that both roots of equation (1) are at infinity.)

Answer 1

Polynomials do not have "roots at infinity." I'm not totally sure about what you mean, but an equation like $f(x)=\frac{1}{x}$ might satisfy your condition. However, polynomials do not have a root at infinity because if such a polynomial $P(x)$ exists, then one factor of $P(x)$ would have to be $(x-\infty)$.

Answer 2

Given a quadratic equation $ax^2+bx+c$ where $a\neq 0$ has roots $\frac{-b}{a}+\sqrt{(\frac{b}{a})^2-\frac{4c}{a}}$ and $\frac{-b}{a}-\sqrt{(\frac{b}{a})^2-\frac{4c}{a}}$. So when $a\neq0$ both roots are finite.

I assume you want to know how to make the roots arbitrary large.

Looking at $\frac{-b}{a}+\sqrt{(\frac{b}{a})^2-\frac{c}{a}}$ you can see what will happen when you choose $|\frac{c}{a}|$ to be arbitrary large.

To be precise see what happens when $\frac{c}{a}=-N$ when $N$ is an natural number.

Your roots will be $\frac{-b}{a}+\sqrt{(\frac{b}{a})^2+4N}$ and $\frac{-b}{a}-\sqrt{(\frac{b}{a})^2+4N}$

Answer 3

In a sense you are re-running the arguments that mathematicians had in the time of Newton and Leibniz, about what you can do with infinitely large and infinitely small quantities.

The conclusion was this:

You can’t put infinity as such into your formulae and expect them to make sense. You can’t put division by zero into your formulae and expect them to make sense.

So we have come up with some firm rules and ways of handling the question.

First of all, though, infinity is a specific concept and not a way of saying “very large”. You must get that clear because otherwise you will find yourself using a language different from everyone else. “Does not fit on any reasonable sheet of paper” is not the same as “Infinite”. The roots of your first equation are not infinite. $2×10^{15}$ is not infinite.

Once you have stopped using “infinity” in a different way from everyone else, you have actually pointed out a very interesting fact.

Basically, what you have found, to take your $A=0$ case, is that the closer to $0$ $A$ becomes, the closer to infinity one of the roots of the quadratic becomes. Intuitively, it is being squeezed out of the real world until when $A=0$ exactly, it has completely gone. Algebraically, you can look at the traditional solution of a quadratic and see how, the smaller that $a$ becomes, one solution (the $(-b-\sqrt{b^2-4ac})/2a$ one) runs off towards infinity because the denominator tends to zero while the numerator doesn’t, while in the other solution (the $(-b+\sqrt{b^2-4ac})/2a$ one) the numerator is also tending to be zero, and although $0/0$ could mean anything or everything or nothing, that is actually a hopeful sign. Some subtle algebraic manoeuvring will eventually reveal that that root tends to $-c/b$.

People run the process backwards as well. Given $Ax^2+Bx+C=0$, where $A$ is a very complicated formula but is also very small, one solves $Bx+C=0$ first and then looks into how that solution is perturbed by the presence of a non-zero $A$. Many advanced calculations in theoretical physics work in this way.

So to conclude:

1. Don’t say “infinite” to mean “very large” or “zero” to mean “very small”.

2. Do think about what happens when something tends to infinity or tends to zero.

3. And congratulations on thinking of this question in the first place. One of the best ways to mathematical discovery is to say to yourself “What a beautiful equation! I wonder how I can break it”.

Answer 4

You can make what your teacher was getting at more precise by using the terminology of sequences and limits . . .

Suppose we have an infinite sequence$\;f_1,f_2,f_3,...\;$of quadratic functions of the form $$f_n = a_nx^2+b_nx + c_n$$ where

• $a_n,b_n,c_n\in\mathbb{R}$, for all $n$.$\\[4pt]$
• Either $a_n > 0$ for all $n$, or $a_n<0$ for all $n$.$\\[4pt]$
• The sequences $(a_n),(b_n),(c_n)$ converge to $A,B,C$ respectively, where $A,B,C\in\mathbb{R}$.$\\[4pt]$
• Each $f_n$ has two distinct real roots, $r_n,s_n$ say, with $r_n < s_n$.

With this setup, your teacher's claims can be rephrased as:

• If $A=0$ and $B\ne 0$, then as $n$ approaches infinity, either $(r_n)$ approaches minus infinity and $(s_n)$ approaches a (finite) real number, or $(s_n)$ approaches infinity and $(r_n)$ approaches a (finite) real number.$\\[6pt]$
• If $A=B=0$ and $C\ne 0$, then as $n$ approaches infinity, $(|r_n|),(|s_n|)$ both approach infinity.

Note that although the sequences $(|r_n|),(|s_n|)$ are allowed to potentially approach infinity or minus infinity, the sequences $(a_n),(b_n),(c_n)$ are assumed to approach finite limits.

In particular, in this context,$\;|C|=\infty\;$is not allowed.

Answer 5

Our maths teacher taught us that conditions for roots of a quadratic equation $ax2+bx+c = 0$ to lie at infinity are ...

What exactly does your teacher mean if he says "roots at infinite"?

If his definition of this expression is that the formula $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\qquad\text{(I)}$$ ... leads to a division by zero, he is wrong!

Why?

I know another form of the same formula which is: $$x_{1,2}=-\frac{b}{2}\pm\sqrt{(\frac{b}{2})^2-c}\qquad\text{(II)}$$

If we apply the second form of this formula to the equation $0x^2+4x+4 = 0$, we get the result(s) $x_{1,2}=-2$.

This solution is wrong ($x=-1$ would be correct), but there is no division by zero.

The reason why we get completely different results in this case if we use forms (I) and (II) of the formula is because assumptions were made when the two forms of the formula were derived:

The assumption made when deriving form (II) was $a=1$. Therefore this form of the formula must not be used when $a\neq 1$.

And as far as I remember the derivation of form (I) of the formula correctly, the derivation of the formula assumes that $a\neq 0$. Therefore the formula must not be used when $a=0$.

I don't see any reason why $0x^2+4x+4 = 0$ should have "roots at infinity" but not have "roots at $-2$". Both statements are simply based on using formulas whose assumptions are not met.

What I want to ask is, are these the only two conditions for roots of a quadratic equation to lie at infinity?

As already said in other comments and answers:

Neither case (A) nor case (B) really have "roots at infinity".

Answer 6

I'm sorry to say that your maths teacher is talking nonsense.

• If $a=0$ and $b\ne 0$, the equation $bx+c=0$ has the single root $x=-c/b$.
• If $a=b=0$, the "equation" $c=0$ is simply true or false, according as $c$ is zero or non-zero. It doesn't have any "roots".

I can only hope that you are simply misremembering what you were told. The roots of a quadratic equation can never "lie at infinity".

By the way, your comment about "so small that they are nearly zero" is also nonsense. Either they are zero or they are not.

Answer 7

The roots to the quadratic equation $y = ax^2+bx+c$ are $$x_1 =\dfrac{-b - \sqrt{b^2-4ac}}{2a} =\dfrac{2c}{-b + \sqrt{b^2-4ac}}$$

and

$$x_2 =\dfrac{-b + \sqrt{b^2-4ac}}{2a} =\dfrac{2c}{-b - \sqrt{b^2-4ac}}$$

If we let $a=0$ and $b > 0$, then the parabola "degenerates" into the line $y=bx+c$, whose root is $x = -\dfrac cb$.

If we look at $\left. \dfrac{-b - \sqrt{b^2-4ac}}{2a}\right|_{a=0}$ we get $x_1 =-\dfrac{-2b}{0}$; which we could interpret as $\infty$.

If we look at $\left. \dfrac{2c}{-b + \sqrt{b^2-4ac}}\right|_{a=0}$ we get $x_2 =-\dfrac cb$.

If we let $a=0$ and $b=0$, then the parabola degenerates into the constant $y=c$.

If $c\ne 0$ then, in the right context, we might say that the roots are $\pm \infty$

If $c = 0$, then the roots are $(-\infty, \infty) = \mathbb R$.