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

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Hello, PF! I'm here with a question. It may be stupid and I may be wrong (I'm surely wrong somewhere in here)...

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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  • #2
hilbert2
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The derivation of the quadratic equation solution formula is not valid if ##a=0## (division by zero), so you can't use ##x=\frac{-b\pm \sqrt{b^2 -4ac}}{2a}##. Do you know how to derive the formula by "completing the square"?
 
  • #3
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The derivation of the quadratic equation solution formula is not valid if ##a=0## (division by zero), so you can't use ##x=\frac{-b\pm \sqrt{b^2 -4ac}}{2a}##. Do you know how to derive the formula by "completing the square"?
Hmm... Looks like I forgot that detail. It actually makes much more sense now.
Now that I think better, the teacher said this back when I studied this.
[itex]ax^2 + bx + c = 0 , \ \ a \not = 0[/itex]

I don't know what do you mean by deriving the formula by "completing the square". Yes, I know how to derive the functions, I learned this year. I don't know what "completing the square" means. I may have learned it, but I may not know the english name.
 
  • #5
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No matter how close [itex]a[/itex] is to zero, as long as it is not precisely zero, the solution formula to the quadratic equation can be used. This means that it makes sense to examine what happens to the zeros in the limit [itex]a\to 0[/itex]. The important tool to this is the Taylor approximation

[tex]
\sqrt{1 + \delta} = 1 + \frac{1}{2}\delta + O(\delta^2)
[/tex]

which holds at the limit [itex]\delta\to 0[/itex]. The solutions to

[tex]
ax^2 + bx + c = 0
[/tex]

can be written as

[tex]
x = \frac{-b \pm |b|\sqrt{1 - \frac{4ac}{b^2}}}{2a}
[/tex]

and now the quantitity [itex]-\frac{4ac}{b^2}[/itex] takes the role of [itex]\delta[/itex]. In the limit [itex]a\to 0[/itex] the zeros can be approximated with formula

[tex]
x = \frac{-b \pm |b|\big(1 - \frac{2ac}{b^2} + O(a^2)\big)}{2a}
[/tex]

If one studies the cases [itex]b>0[/itex] and [itex]b<0[/itex] separately, one finds that eventually the zeros can be simplified to the following forms:

[tex]
x = -\frac{c}{b} + O(a)\quad\textrm{or}\quad x = -\frac{b}{a} + \frac{c}{b} + O(a)
[/tex]

Thus we see that one of the zeros approaches the number [itex]-\frac{c}{b}[/itex], which is the solution to

[tex]
bx + c = 0
[/tex]

while the other zero diverges to [itex]\pm\infty[/itex].
 
  • #6
jbunniii
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Hello, PF! I'm here with a question. It may be stupid and I may be wrong (I'm surely wrong somewhere in here)...

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.
No, the degree of a (nonzero) polynomial is by definition the highest power which occurs with a nonzero coefficient. So ##0x^2 + ax + b## has degree 1, not 2.
 

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