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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.)

enter image description here

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closed as unclear what you're asking by Matthew Towers, user21820, YiFan, Xander Henderson, Toby Mak Jul 14 at 6:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Neither of your examples have "roots at infinity". $\endgroup$ – Lord Shark the Unknown Jul 7 at 5:38
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    $\begingroup$ Well the roots of (1) are $x=\pm 2\times 10^{15}$ which are certainly not "at infinity". $\endgroup$ – Lord Shark the Unknown Jul 7 at 5:50
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    $\begingroup$ "A very large number" is not infinite, and hence a root at one is not a root at the the other. If a=0, it isn't even a quadratic and if b=0 there is no equation in $x$, so there's no point discussing roots at all. If you're talking about the limit as the coefficients close to zero, then you can only conclude the limit. "How do I explain these discrepancies?" - by acknowledging your conception of how roots or limits work is erroneous, and asking other questions to correct it. $\endgroup$ – Nij Jul 7 at 5:50
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    $\begingroup$ @πtimese $-10^{100000000000000}$ is not an "infinitely large negative number". $\endgroup$ – Lord Shark the Unknown Jul 7 at 6:01
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    $\begingroup$ Even if it is a googol to the power of Graham's number, it is still closer to $0$ than $\infty$. $\endgroup$ – BadAtGeometry Jul 7 at 6:03
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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”.

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    $\begingroup$ Applying a binomial formula to get $$\frac{-b+\sqrt{b^2-4ac}}{2a}=-\frac{2c}{b+\sqrt{b^2-4ac}}$$ is not exactly "subtle" and also not a very long calculation to get $-\frac cb$ for $a\to0$. // Note that the validity of the assignment to "finite" and "infinite" root depends on $b$ being positive. $\endgroup$ – LutzL Jul 7 at 7:44
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    $\begingroup$ Thanks a lot for this thorough explanation. I understand it now. As you’ve pointed out, I thought infinity referred to an extremely large number, that’s where I was tripping up. Thanks for clearing it up, and thanks again for the beautiful explanation $\endgroup$ – π times e Jul 7 at 12:48
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    $\begingroup$ What a great answer to a question that had me (and I'm sure many others) tempted to just write it off as confused nonsense. $\endgroup$ – dbmag9 Jul 7 at 21:30
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    $\begingroup$ @dbmag9 let's try to remain cordial here. A question can appear to be nonsense to one person and not-nonsense to another, but "confused" applies to a real person (in this case the OP) and in this case they seem to have made a genuine effort to understand something, formulate, compose, and post a question. $\endgroup$ – uhoh Jul 7 at 23:03
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    $\begingroup$ @uhoh My intention with 'confused' was to apply to the text rather than its author (and in any case I don't think a person being confused means they haven't made a genuine effort), but I guess that didn't come through – apologies, OP, if I offended! $\endgroup$ – dbmag9 Jul 8 at 11:47
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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.

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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 $(x-\infty)$.

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    $\begingroup$ @πtimese You can say whatever you like; that doesn't stop what you are saying from being false. $\endgroup$ – Lord Shark the Unknown Jul 7 at 6:00
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    $\begingroup$ I’m not just saying whatever I like, I’m trying to understand it, or trying to clear up my doubts. I think I’m supposed to ask questions if I don’t understand. Not sure why you are after me $\endgroup$ – π times e Jul 7 at 6:01
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    $\begingroup$ For a start, you keep referring to specific finite numbers as "infinitely large". Why? $\endgroup$ – Lord Shark the Unknown Jul 7 at 6:03
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    $\begingroup$ "infinity is a notion, not a number", "an infinitely large positive number". You seem to be contradicting yourself. $\endgroup$ – Lord Shark the Unknown Jul 7 at 6:06
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    $\begingroup$ @πtimese If c is actually $\infty$, then the graph will never touch the x-axis. $\endgroup$ – BadAtGeometry Jul 7 at 6:09
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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".

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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 $\frac{-b}{a}-\sqrt{(\frac{b}{a})^2+4N}$

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  • $\begingroup$ Your root formula is missing a factor $\frac12$. Also, what is the conclusion you arrive at? How does one root compensate the large terms to stay finite? $\endgroup$ – LutzL Jul 7 at 7:47
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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 $(-\infty, \infty) = \mathbb R$.

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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.

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    $\begingroup$ -1 for two gratuitous instances of "nonsense" (though not double nonsense). I'm sure there is a way to write this answer without them. $\endgroup$ – uhoh Jul 7 at 23:07
  • $\begingroup$ @uhoh: I suppose I must take it as a compliment that you remember my answers! And I assure you that this answer couldn't have been written without using the word "nonsense". $\endgroup$ – TonyK Jul 7 at 23:29
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    $\begingroup$ "your maths teacher is talking nonsense" and "By the way, your comment... is also nonsense." are necessary? You don't see them as condescending and potentially hurtful? $\endgroup$ – uhoh Jul 7 at 23:42

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