39
$\begingroup$

It is commonly known that $ax^2+bx+c=0$ have two solutions $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ but how to prove that another root couldn't exist?

I think derivation of quadratic formula is not enough....

$\endgroup$
  • 15
    $\begingroup$ It's referred to as "The fundamental theorem of algebra" that counting multiple roots and complex roots, polynomials of degree n have n roots. You can google it. $\endgroup$ – fleablood Dec 8 '16 at 17:36
  • 12
    $\begingroup$ "given the tag, I'm assuming we're working in R" and "Alternatively, you could note that the derivative has exactly one zero and thus by Rolle's theorem"..... Um...? $\endgroup$ – fleablood Dec 8 '16 at 17:41
  • 20
    $\begingroup$ @MathematicsStudent1122 - pre-calculus students who have reached the point of quadratic equations and have heard that quadratics have two roots have likely encountered complex numbers. However, since they are pre-calculus, they are highly unlikely to be familiar with derivatives and Rolle's theorem. $\endgroup$ – Paul Sinclair Dec 9 '16 at 2:18
  • 8
    $\begingroup$ @AtulMishra - Actually, derivation of the quadratic formula is enough: That derivation starts with an arbitrary solution to the polynomial, and shows that it has one of those two values. Therefore those two values are the only ones possible. $\endgroup$ – Paul Sinclair Dec 9 '16 at 2:20
  • 5
    $\begingroup$ Can you assume that all polynomials can be written in the form $c(x-x_1)(x-x_2)\ldots$? Or do you need a proof of that too? Because if you can assume that, then the answer is trivially easy (just observer that having more than 2 roots implies a degree for $x$ greater than 2.) $\endgroup$ – Mehrdad Dec 9 '16 at 9:55

11 Answers 11

172
$\begingroup$

Suppose there are three distinct roots $x,y,z$. One has $$\begin{cases}ax^2+bx+c=0\\ay^2+by+c=0\\az^2+bz+c=0\end{cases}\Rightarrow\begin{cases}a(x^2-y^2)+b(x-y)=0\\a(x^2-z^2)+b(x-z)=0\end{cases}\Rightarrow\begin{cases}a(x+y)+b=0\\a(x+z)+b=0\end{cases}$$ It follows $$a(z-y)=0\Rightarrow z=y$$ which is a contradiction.

$\endgroup$
  • 6
    $\begingroup$ your last system should be $a(x+y)+b = 0$ and $a(x+z)+b=0$. The rest stand, great proof. $\endgroup$ – Alain Remillard Dec 8 '16 at 18:30
  • 5
    $\begingroup$ I often mistaken for hurry (many times, because of my bad English, when I have finished my writing an equal answer has already been published.) Thank you. $\endgroup$ – Piquito Dec 9 '16 at 0:43
  • 8
    $\begingroup$ @Wildcard Right. And it's also implicit that it's ok to divide by these terms because by the assumption of distinct roots neither of these can be zero. $\endgroup$ – Hugh Meyers Dec 9 '16 at 7:41
  • 31
    $\begingroup$ It would be great if we had an $a \ne 0$ in there. $\endgroup$ – Daniel R. Collins Dec 9 '16 at 15:09
  • 3
    $\begingroup$ @DanielR.Collins - Recall the phrase "quadratic equation" in the title of the OP's posting. If $a=0$, one doesn't have a quadratic equation to start with. Thus, the condition $a\ne0$ is true automatically as long as one is working with (non-trivial!!) quadratic equations. $\endgroup$ – Mico Dec 11 '16 at 9:21
51
$\begingroup$

I think derivation of quadratic formula is not enough....

Yes it is. The derivation is of the form if $ax^2+bx+c=0$, then $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. The derivation is a proof if you pay attention.

The trickiest step is simply that if $y^2 = k$ for $k \geq 0$ then $y = \pm \sqrt k$, if you do not take this as evident.

$\endgroup$
  • 6
    $\begingroup$ All that needed to be said. $\endgroup$ – Jack M Dec 9 '16 at 19:01
  • 1
    $\begingroup$ @JackM thanks. I do think the answers regarding polynomial factorization over fields/integral domains, and how this relates to the fundamental theorem of algebra, are pretty useful. $\endgroup$ – djechlin Dec 9 '16 at 22:10
27
$\begingroup$

$$0 = ax^2 + bx + c$$

We solve this equation by completing the square. It offers up to two distinct solutions. The name we give to the general solutions is the quadratic formula. That's all there is.

If we consider the case of real solutions, and you think there may be a sneaky third solution, remember that $f(x) = ax^2 +bx +c$ can be plotted as below (depending on the sign of $a$). How many times could a parabola cross a horizontal line?

parabola intersecting various horizontal lines

$\endgroup$
  • 2
    $\begingroup$ this is but a specific degenerate case of 2D plane! Generalize it at very least to 3D complex space and you would easily see there are much more solutions !!! :-D $\endgroup$ – Arioch 'The Dec 9 '16 at 12:00
  • 10
    $\begingroup$ @BillDubuque Yes, that is all true. What's also true is that Math.SE exists for users of all levels. It's beneficial to provide responses that could meet varying expectations, unless the problem specifies. $\endgroup$ – zahbaz Dec 9 '16 at 18:33
  • 5
    $\begingroup$ This feels circular to me, don't you need to prove that all quadratics have a graph like that? $\endgroup$ – RemcoGerlich Dec 9 '16 at 19:31
  • 4
    $\begingroup$ Useful <> Rigorous, it is possible for ideas that lack rigor to be useful, and moreover, useful to mathematical understanding. The point is, is it likely to be an accurate model? If so, its' potentially useful as an aid to understanding, regardless of whether or not we have a proof of its accuracy $\endgroup$ – Brad Thomas Dec 9 '16 at 20:21
  • 4
    $\begingroup$ @RemcoGerlich standupmaths made a "There is only One True Parabola" video demonstrating all parabolas are similar. $\endgroup$ – Nick T Dec 9 '16 at 21:23
17
$\begingroup$

A more general answer to this question lies in the following theorem:

Theorem If $P(x)$ is a polynomial of degree $n$, and $a$ is a value for which $P(a) = 0$, then $P(x) = (x - a)Q(x)$, where $Q(x)$ is a polynomial of degree $n - 1$.

This theorem is a simple consequence of polynomial long division. By long division, $P(x) = (x - a)Q(x) + R(x)$, for some polynomials $Q(x), R(x)$ with the degree of $R(x)$ less than the degree of $(x-a)$. But since $x - a$ is of degree 1, that means $R(x)$ is of degree $0$. I.e., $R(x) = R$, a constant.

But $P(a) = 0$, so $0 = (a - a)Q(a) + R$, and so $R = 0$ and we get just $P(x) = (x-a)Q(x)$. Since the degree of the product of two polynomials is the sum of their degrees, the degree of $P(x)$ is one greater than that of $Q(x)$, so the degree of $Q(x)$ must be $n-1$.


Now, if $P_n(x)$ is of degree $n > 0$ and $a_n$ is a root, then $$P_n(x) = (x - a_n)P_{n-1}(x)$$ for some $n-1$ degree polynomial $P_{n-1}(x)$. If $P_n(x)$ has another root $a \ne a_n$, then $a$ must also be a root of $P_{n-1}(x)$:

$$0 = P_n(a) = (a - a_n)P_{n-1}(a)$$ Since $a - a_n \ne 0$, we can divide it out to get $P_{n-1}(a) = 0$.

Conversely, if $a_{n-1}$ is a root of $P_{n-1}$, then $$P_n(a_{n-1}) = (a_{n-1} - a_n)P_{n-1}(a_{n-1}) = 0$$ So $a_{n-1}$ must also be a root of $P_n$ (which may be the same or different from $a_n$). We can also apply the theorem to $P_{n-1}$ and $a_{n-1}$: $$P_{n-1}(x) = (x - a_{n-1})P_{n-2}(x)$$ for some degree $n-2$ polynomial $P_{n-2}(x)$. By combining, we see that $$P_n(x) = (x - a_n)(x - a_{n-1})P_{n-2}(x)$$ As long as we can keep finding roots for the reduced polynomials, we can keep this up. If we can find $k$ such roots, $$P_n(x) = (x - a_n)(x - a_{n-1})(x - a_{n-2})...(x - a_{n+1-k})P_{n-k}(x)$$ Then $P_{n-k}(x)$ has to be a polynomial of degree $n-k$.

If we can find $n$ such roots, then $$P_n(x) = (x-a_n)(x-a_{n-1})...(x-a_1)P_0$$ where $P_0$ is a constant (a $0$-degree polynomial). $P_0 \ne 0$, since if it were we would have $P_n(x) = 0$ everywhere. But then the degree of $P_n$ would be $0$ (or less - some people define the degree of the $0$ to be $-\infty$), contrary to our original condition on $P_n(x)$. So in this case, $P_n(x)$ cannot have any other roots distinct from $a_1, a_2, ..., a_n$, since any other value would leave all factors in the expression non-zero.

So $P_n(x)$ can have at most $n$ roots.

The Fundamental Theorem of Algebra says that any non-constant polynomial over the complex numbers has a root. This theorem requires a substantial development of the properties of complex numbers to prove. But by it, we see that the process above does not terminate until you get to the constant. Thus a polynomial of degree $n$ will always have exactly $n$ roots $a_1, a_2, ..., a_n$. But remember that the $a_i$ values do not have to be distinct. The number of times a particular value occurs in this list is called the multiplicity of the root. So you only get $n$ if you count the roots by their multiplicity.

$\endgroup$
16
$\begingroup$

Hint $\ $ Suppose $\,f(x)\,$ is a polynomial of degree $2$ with coefficients in a field $\,F$ (e.g. $\Bbb Q,\Bbb R,\Bbb C)$ and suppose that $\,f\,$ has $\,2\,$ distinct roots $\,a,b.\,$ By the Bifactor Theorem below we deduce that $\,f(x) = c(x\!-\!a)(x\!-\!b)\,$ for $\,c\in F.\,$ Thus if $\,d\neq a,b\,$ then $\,f(d) = c(d\!-\!a)(d\!-\!b)\ne 0\,$ since each factor is $\ne 0\,$ (recall $\,x,y\ne 0\,\Rightarrow\,xy\ne 0\,$ in a field). So $\,f\,$ has at most $2$ distinct roots.

Bifactor Theorem $\ $ Suppose that $\rm\,a,b\,$ are elements of a field $\rm\,F\,$ and $\rm\:f\in F[x],\,$ i.e. $\rm\,f\,$ is a polynomial with coefficients in $\rm\,F.\,$ If $\rm\ \color{#C00}{a\ne b}\ $ are elements of $\rm\,F\,$ then

$$\rm f(a) = 0 = f(b)\ \iff\ f\, =\, (x\!-\!a)(x\!-\!b)\ h\ \ for\ \ some\ \ h\in F[x]$$

Proof $\,\ (\Leftarrow)\,$ clear. $\ (\Rightarrow)\ $ Applying Factor Theorem twice, while canceling $\rm\: \color{#C00}{a\!-\!b\ne 0},$

$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in F[x]\\ \rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in F[x]\\ &\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$

Remark $ $ More generally, by iterating the Factor Theorem (as above) we can show that a nonzero polynomial $\,f\,$ over a field (or domain) has no more roots than its degree $\,n.\,$ Indeed if $\,f\,$ has $\,\ge n\,$ distinct roots $\,r_i$ then inductively applying the Factor Theorem shows $\,f = c(x\!-\!r_1)\cdots (x\!-\!r_n),\,$ so $\ r\ne r_i\Rightarrow\, f(r)= c(r\!-\!r_1)\cdots (r\!-\!r_n) \ne 0\,$ by all factors are $\ne 0.\,$ So $\,f\,$ has at most $\,n\,$ roots.

The above root-bound property completely characterizes integral domains: the ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in $\rm\:D.\:$ For the simple proof see my post here, where I illustrate it constructively in $\rm\: \mathbb Z/m\: $ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd.\,$

The quadratic case of this result is at the heart of some integer factorization algorithms, which e.g. attempt to factor $\rm\:m\:$ by searching for a square-root of $1$ that is nontrivial $(\not\equiv \pm1)$ in $\rm\: \mathbb Z/m.$

$\endgroup$
  • $\begingroup$ See this answer for a generalization from domains to any commutative ring. $\endgroup$ – Bill Dubuque Jul 14 '17 at 14:14
13
$\begingroup$

Let $$r=\frac{-b+\sqrt{b^2-4ac}}{2a},\quad s=\frac{-b-\sqrt{b^2-4ac}}{2a}.$$ Simple calculation shows that $$r+s=-\frac ba\quad\text{ and }\quad rs=\frac{b^2-(b^2-4ac)}{4a^2}=\frac ca.$$ Thus $$a(x-r)(x-s)=a[x^2-(r+s)x+rs]=ax^2+bx+c.$$ If $t$ is any root of the quadratic equation $ax^2+bx+c=0,$ then $$a(t-r)(t-s)=at^2+bt+c=0.$$ Since $a\ne0$ this means that $$(t-r)(t-s)=0$$ whence $$t-r=0\quad\text{ or }\quad t-s=0,$$ i.e., $$t=r\quad\text{ or }\quad t=s.$$

$\endgroup$
10
$\begingroup$

Suppose it has three roots and $a\neq0$.

The hypotheses of Rolles Theorem are satisfied then there will exist two roots of the derivative and a root of the second derivative which is a constant $(=2a)$.

$\endgroup$
5
$\begingroup$

I think I can simplify the 'polynomial long division' answer. The special case of polynomial long division says that, for any polynomial $P$, and any real number $a$, $$P(x) = Q(x)(x-a) + R$$ for some polynomial $Q$ and constant $R$ (use long division to divide $P$ by $x-a$ and observer that $R$ is required to be a constant since it must have lower degree than $x-a$, which is a first-degree polynomial).

The above is true for all $x$, so substituting $x=a$ we get $$P(a) = Q(a)(a-a) + R$$ Obviously, $a-a=0$, so $R=P(a)$.

If $a$ is a 0 of $P$ ($P(a)=0$), then $R=0$, so $x-a$ divides $P(x)$.

Now, if we have any distinct root $a$ of a quadratic polynomial $P$, we know $$P(x) = Q(x)(x-a)$$ $Q$ must be a first-degree polynomial, since anything higher-degree, multiplied by a first-degree polynomial, would produce a higher-than-second-degree polynomial. So $$P(x) = (x-b)(x-a)$$ Now, assuming we're working over an integral domain (which $\mathbb R$ and $\mathbb C$ both are), $$P(x) = 0 \Rightarrow x-b=0 \text{ or } x-a=0$$ So $a$ and $b$ are the only zeros of $P$ (although it is possible that $a=b$).

$\endgroup$
5
$\begingroup$

Suppose not. Then there are at least 3 roots and so $$P(x)=(x-x_1)(x-x_2)(x-x_3)Q(x)$$ This is at least a cubic so this cannot happen.

$\endgroup$
1
$\begingroup$

If you wrote the theorem out, it would look like this

THEOREM: Let $a,b,$ and $c$ be real numbers with $a \ne 0$. Then $ax^2+bx+c=0$ if and only if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

That means

if $ax^2+bx+c=0$, then $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

and

if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, then $ax^2+bx+c=0$.

$\endgroup$
-7
$\begingroup$

I don't know whether topic starter asked this question to troll us or not, but the answer is quite complicated.

This problem is known as ‘Fundamental Theorem of Algebra’: number of roots (including complex ones and multiple ones) of a polynom is equal to the polynom's degree. Proving this theorem is quite a sophisticated thing since you can't prove in real numbers — only complex numbers theory introduced strict proof for any n. See: https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

The implication of that theorem is that no polynom could have more roots than it's degree, so quadratic equation couldn't have more than two solutions.

$\endgroup$
  • 4
    $\begingroup$ No, FTA is existence of a root, which is only true in the complex numbers, and requires complex analysis. At most deg(poly) roots is true in any R[x] for R an integral domain. This is extensively covered in the other answers already, btw. $\endgroup$ – djechlin Dec 9 '16 at 15:29
  • 1
    $\begingroup$ And even then, why would wondering why the FTA is true be a troll? Why wouldn't you wonder why it's true? $\endgroup$ – djechlin Dec 9 '16 at 15:33

protected by user223391 Dec 11 '16 at 0:36

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.