How can I prove that all quadratic equations are not injective? I was trying to prove that any quadratic formula ($ax^2 + bx + c$) will not be injective, but I have a little problem.
I started by assuming $f(x) = f(y)$.
We can put x and y into the general form of quadratic function, and we get
$ax^2 + bx + c = ay^2 + by + c$
Subtract c from both sides and organise a little bit and you get
$a(x^2-y^2) + b(x-y)=0$
$a(x+y)(x-y) + b(x-y)=0$
Here we can assume that $x\ne y$, hence $x-y\ne 0$, so divide both sides by x-y, and we get
$a(x+y)+b=0$
However, from here I couldn't find any contradictions, which is a problem because there must be a contradiction as quadratic functions are not injective. Can anyone tell me what the contradiction is here?
 A: There is no contradiction. You have in fact come upon the condition such that $f(x)=f(y)$ for $x\neq y$. With a little rearranging, you have
$$x=-\frac{(b+ay)}{a}$$
(if $a=0$ then you don't have a quadratic). In fact, this works both ways which can be shown by
$$f(y)-f\left(-\frac{(b+ay)}{a}\right)=ay^2+by+c-a\left(-\frac{(b+ay)}{a}\right)^2-b\left(-\frac{(b+ay)}{a}\right)-c$$
$$=ay^2+by-\frac{b^2}{a}-2by-ay^2+\frac{b^2}{a}+by=0$$
is true for all $y$.
A: This is good stuff so far. You now know that you get $x = y$ if you have $a(x + y) + b \neq 0$, in which case, there is no contradiction. So, let's focus on when $a(x + y) + b = 0$. Can you find two values of $x$ and $y$, that are not equal to each other, but for which $a(x + y) + b = 0$? Or, equivalently, $x + y = -\frac{b}{a}$?
There are many possible answers to this, but here's one: we can let $x = -\frac{b}{2a} + 1$ and $y = -\frac{b}{2a} - 1$. These values of $x$ and $y$ could well contradict the definition of injectivity. We have
\begin{align*}
&a\left(-\frac{b}{2a} + 1\right)^2 + b\left(-\frac{b}{2a} + 1\right) + c \\
= \, &a\left(\frac{b^2}{4a} - 2\frac{b}{2a} + 1\right) - \frac{b^2}{2a} + b + c \\
= \, &\frac{ab^2}{4a} - \frac{2ab}{2a} + a - \frac{b^2}{2a} + b + c \\
= \, &\frac{b^2}{4} + a - \frac{b^2}{2a} + c.
\end{align*}
Try substituting in $-\frac{b}{2a} - 1$ instead, and show that it comes to the same value.
A: One way to tackle it is by resorting to their graphs. How does the graph of a quadratic look like? Use the horizontal line test. (Should be good enough a hint.)
A: You just found the value of $x$ and $y$ such that $f(x) = f(y)$ when $x \ne y$.
To see it in a constructuve way observe that $f(x) = ax^2 + bx + c = \frac{1}{4a}\left( (2ax +b)^2 + 4ac - b^2 \right)$
Now observe that at $y = -x - b/a$ we have $2ay + b = -2ax -b$ so it follows that $f(x) = f(y)$.
A: Let us assume the quadratic functions are injective which implies $f(x)=f(y)\implies x=y$
Here $a,b,c$ are some constants and $a\neq 0$
As per you $f(x)=f(y) $ and $x\neq y\implies x+y=-\frac{b}{a} $ which is equivalent to
$f(x)=f(y)\implies x=y $ or $x+y=-\frac{b}{a}$
The above statements are contradiction showing that our assumption is wrong.
A: Alternative approach:
The formula for computing the roots of $~~ax^2 + bx + c = 0~~$ is $$x = \frac{1}{2a}\left[-b \pm \sqrt{b^2 - 4ac}\right].\tag 1$$
Based on (1) above, in Real Analysis, the attempt to solve $~~f(x) = ax^2 + bx + c = 0~~$ will result in $3$ possibilities:
$\underline{\text{Case 1}}: ~b^2 - 4ac < 0.$ 
In this case, there will be no real roots.  Since the issue is whether the computation of the root(s) of $f(x)$ is injective rather than surjective, it seems reasonable to presume that in this case, injectivity has not been violated.
$\underline{\text{Case 2}}: ~b^2 - 4ac = 0.$ 
In this case, there will be one root, which is repeated.  Therefore, the association between $f(x)$ and its (single) root may again be regarded injective.
$\underline{\text{Case 3}}: ~b^2 - 4ac > 0.$ 
In this case, there will be two distinct roots, so the association between $f(x)$ and its (two) roots will not  be injective.
