How can I prove that all quadratic equations are not injective?I was trying to prove that any quadratic formula (ax2+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 getax2+bx+c=ay2+by+cSubtract c from both sides and organise a little bit and you geta(x2−y2)+b(x−y)=0a(x+y)(x−y)+b(x−y)=0Here we can assume that x≠y, hence x−y≠0, so divide both sides by x-y, and we geta(x+y)+b=0However, from here I couldn't find any contradictions, which is a problem because there must be a contradiction as quadratic functions are not injective.

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How can I prove that all quadratic equations are not injective?I was trying to prove that any quadratic formula (ax2+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 getax2+bx+c=ay2+by+cSubtract c from both sides and organise a little bit and you geta(x2−y2)+b(x−y)=0a(x+y)(x−y)+b(x−y)=0Here we can assume that x≠y, hence x−y≠0, so divide both sides by x-y, and we geta(x+y)+b=0However, from here I couldn&#039;t find any contradictions, which is a problem because there must be a contradiction as quadratic functions are not injective.

Mason Knight · Accepted Answer

There is no contradiction. You have in fact come upon the condition such that f(x)=f(y) for x≠y. With a little rearranging, you havex=−(b+ay)a(if a=0 then you don&#039;t have a quadratic). In fact, this works both ways which can be shown byf(y)−f(−(b+ay)a)=ay2+by+c−a(−(b+ay)a)2−b(−(b+ay)a)−c=ay2+by−b2a−2by−ay2+b2a+by=0is true for all y.

Theodore Davila · Accepted Answer

You just found the value of x and y such that f(x)=f(y) when x≠y.To see it in a constructuve way observe that f(x)=ax2+bx+c=14a((2ax+b)2+4ac−b2)Now observe that at y=−x−ba we have 2ay+b=−2ax−b so it follows that f(x)=f(y).

How can I prove that all quadratic equations

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