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Proof:(By Contradiction)

On the contrary, a quadratic equation can have more than two roots. Suppose $\alpha$, $\beta$ and $\gamma$ are three distinct roots of a quadratic equation. Then these three roots must satisfy the quadratic equation. So, we can write it as the following:

$ a\alpha^{2}+b\alpha+c=0 ...(1)$

$a\beta^{2}+b\beta+c=0 ...(2)$

$a\gamma^{2}+b\gamma+c=0 ...(3)$

$(1)-(2):$ we have

$a(\alpha^{2}-\beta^{2})+b(\alpha-\beta)=0$

$(\alpha-\beta)\{a(\alpha+\beta)+b\}=0$

Since $\alpha-\beta≠0$. Therefore, $a(\alpha+\beta)+b=0 ...(4)$

$(2)-(3):$ we have

$a(\beta^{2}-\gamma^{2})+b(\beta-\gamma)=0$

$(\beta-\gamma)\{a(\beta+\gamma)+b\}=0$

Since $\beta-\gamma≠0$. Therefore, $a(\beta+\gamma)+b=0 ...(5)$

$(4)-(5):$ we have $a(\alpha-\gamma)=0$ Since $a≠0$. Therefore, $\alpha-\gamma=0 \implies \alpha=\gamma$. This is a contradiction to that fact that $\alpha$, $\beta$ and $\gamma$ are distinct. This establishes the proof.

Is this correct approach to prove it?

Waseem F-Name Ali Nawaz
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    Might I suggest avoiding using Latin and Greek letters (and corresponding ones at that!) in the same problem? It's a tad confusing haha. Edit: this looks correct though! – Cameron L. Williams Jun 08 '21 at 19:07
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    Well.. I think this is correct. But there is a much more straightforward argument. If $\alpha,\beta$ and $\gamma$ are three different solutions, your polynomial would be divisible by $(x-\alpha), (x-\beta)$ and $(x-\gamma)$ so it would have the form $(x-\alpha)(x-\beta)(x-\gamma)q(x)$ for some polynomial $q$. But this is impossible, since the latter has degree $\ge 3$. – GReyes Jun 08 '21 at 19:08
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    I would take this approach as well @GReyes but if you strictly want to limit yourself to zeroes (neglecting the algebra behind polynomials), then OP's approach is fine. It depends on what machinery you can use or want to use. – Cameron L. Williams Jun 08 '21 at 19:11
  • Yes, that works. One way that doesn't tie you to degree $2$ is to note that the matrix of the system of equations that you got is a Vandermonde matrix corresponding to the values $\alpha,\beta,\gamma$. The determinant of a Vandermonde is the product of all difference of those numbers. Since $a\neq0$ then the system (thinking of $a,b,c$ as the unknowns) has a non-trivial solution. Therefore, the Vandermonde determinant would have to be zero. As a consequence, at least one of the differences of the $\alpha,\beta,\gamma$ should be zero. – plop Jun 08 '21 at 19:12
  • @CameronWilliams Sure. I agree that it depends on the context. – GReyes Jun 08 '21 at 19:19
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    @GReyes that assumes you know that if $r$ is a root of a polynomial then $x-r$ factors the polynomial. Not hard to prove but.... if we could take that for granted then we could simply say we should take it for granted that a polynomial of degree $k$ has at most $k$ roots... which must of us do take for granted.... Also to say $x-\alpha, x-\beta, x-\gamma$ divide $P(x)$ implies the product $(x-\apha)(x-\beta)(x-\gamma)$ implies that we somehow know that $(x-\alpha),(x-\beta),(x-\gamma)$ are somehow "relatively prime"... which they are but that may be beyond the scope of what is known yet. – fleablood Jun 08 '21 at 19:27
  • You must add that your equation is with coefficients in $\mathbb C$ because, for example, $x^2-1=0$ have more than two roots modulo $16$. – Ataulfo Jun 08 '21 at 19:27
  • Alternatively, complete the square, then the problem reduces to showing that $x^2+c=0$ has $0$, $1$, or $2$ (real) roots depending on the sign of $c$. – dxiv Jun 08 '21 at 20:00
  • related – C Squared Jun 08 '21 at 20:10

1 Answers1

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The comments immediately following the question discuss the OP's analysis, which I (also) consider valid. I favor the following alternative approach:

Suppose $f: \Bbb{R} \to \Bbb{R}$ has form
$a_n x^n + a_{n-1}x^{n-1} + \cdots a_1 x^1 + a_0 ~: ~a_n \neq 0.$
That is, $f(x)$ is an $n$-th degree polynomial.

Suppose further that $f(r) = 0$.

Then, there exists a polynomial $g(x)$ of degree $(n-1)$ such that $f(x) = (x-r)g(x).$

Proof

For $s \in \Bbb{Z^+}, ~~x^s - r^s = $
$(x - r) \times (x^{s-1} + x^{s-2}r + x^{s-3}r^2 + \cdots + xr^{s-2} + r^{s-1})$.

Let $h(s,r)$ denote the polynomial of degree $(s-1)$ that represents $\dfrac{x^s - r^s}{x - r}.$

Since $f(r) = 0,$ you have that
$\displaystyle f(x) = f(x) - f(r) = \sum_{k=1}^n \left[(x - r) \times a_k \times h(k,r)\right].$

Since each term has the common factor of $(x - r)$, this implies that $f(x) = f(x) - f(r) = $
$\displaystyle (x - r) \left\{\sum_{k=1}^n \left[a_k \times h(k,r)\right]\right\}.$

In the above summation, only the term that represents $(k = n)$ will feature a polynomial of degree $(n-1)$. Therefore, the overall summation must represent a polynomial $g(x)$ of degree $(n-1)$.

Therefore $f(x) = (x-r)g(x).$

user2661923
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