Hello, PF! I'm here with a question. It may be stupid and I may be wrong (I'm surely wrong somewhere in here)...(adsbygoogle = window.adsbygoogle || []).push({});

Okay, so we have:

[itex]f(x) = 0x^2+ax+b = ax+b[/itex]

You agree with me, right? But what if I won't remove [itex]0x^2[/itex] from my equation? That would leave it like [itex]f(x) = 0x^2 + ax + b[/itex] which is still the same thing but it is a 2nd degree function.

So if I try to solve [itex] f(x) = 0 [/itex] and treat it like a 2nd degree function, it would look like this:

[itex]0x^2 + ax + b = 0;[/itex]

[itex]Δ = a^2 - 4*0*b = a^2 ≥ 0[/itex]

So.. [itex]∃ x_1 , x_2 , \in\mathbb R[/itex]

[itex]x_1,_2 = \frac{-a\ \ \pm\ \ a}{0}[/itex]

And that leaves us with the solutions:

[itex]x_1 = \frac{-2a}{0} \not\in\mathbb R[/itex]

and

[itex]x_2 = \frac{0}{0} \not\in\mathbb R[/itex]

However, we all know that a linear function:

[itex]g(x) = ax+b[/itex] has the solution: [itex]x = \frac{-b}{a}[/itex]

So why do I get such different results if I treat the same function differently? I mean, did I think something wrong in my reasoning? It's not part of my homework or anything, it just came to me while reading a topic here and I had to write it down to see what would happen... So it got my attention. Something must be wrong here, but teoretically [itex]f(x) = g(x)[/itex] ... So I got curious about it.

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# F(x) = 0x^2 + ax + b

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