Can you find the zeros of a quadratic equation using differentiation?

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Given an equation in the form 0 = ax2 + bx + c, why is it not possible to simply differentiate both sides, and produce 0 = 2ax + b, and solve for x, instead of going through the drudgery of the quadratic equation?

My knowledge of calculus tells me that it is because 0 = ax2 + bx + c is not a function, and is only fulfilled by 2 real values, but I want there to be deeper reason. =|
 

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  • #2
Hurkyl
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There is no deep reason; it's an entirely straightforward thing.

If you're taking 0 = ax2 + bx + c as a hypothesis, then x is most certainly definitely not an indeterminate variable ranging over the real line; differentiation doesn't make sense*.

However, if you start with the hypothesis that x is an indeterminate variable ranging over the real line, then 0 = ax2 + bx + c is simply not an identity it satisfies.



Furthermore, even if differentiation did make sense, you've made the common mistake of reading your derivation backwards. In general if you start with an equation P and do some manipulations to derive an equation Q, and find some solution to Q, it does not imply that it is also a solution of P.



*: at least, non-trivially. In this situation, it would make sense to say "the derivative of anything is zero". And, you can check, that differentiating both sides of the equation gives another identity, albeit a useless one.
 
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Thanks for clarifying,

How awesome it would be if that worked.

I blame my physics homework for making me realize how awesome that would be, and then realizing that it doesnt work.

=[

Thanks again -.-
 
  • #4
HallsofIvy
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I have always thought that it would be awsome if "x= 1" were the correct answer to every problem but that doesn't work either!
 
  • #5
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The derivative of a quadratic could be represented with a linear function, whose root represents the max or min value of that quadratic... The derivative would only represent the instantaneous slopes of the quadratic, so at the roots the slope could be anything... So the derivative wouldn't be able to indicate the roots in any way
 
  • #6
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Have you tried solving 0 = 2ax + b for x? I bet you'll find the position of the solution on y=ax2+bx+c to be quite unexpected.
 
  • #7
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Have you tried solving 0 = 2ax + b for x? I bet you'll find the position of the solution on y=ax2+bx+c to be quite unexpected.

Solving for x on the derivative function only gives you the "x" value for the vertex of the original quadratic.

eg. 0 = x2 - 13x + 40
x = -b/2a = -(-13)/(2(1)) = 13/2

Derivative: 0 = 2x - 13
x = 13/2 (the "x" portion of the vertex)

and the roots of the original function are 5 and 8 (so they weren't found by solving for x)
 
  • #8
Char. Limit
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Of course, the vertex IS exactly the midpoint of the two zeros (if there are any), so that's something at least. You could probably use that to show that if there are two non-real roots, then they are complex conjugates (how else would their midpoint be real?)
 
  • #9
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When you start with an equation, where both sides are a differentiable function, it is true that the derivative of the left is equal to the derivative of the right.

However, the act of differentiation has changed what the solution to that equation would mean.

Edit; one reason you can't get the zeros is because it is the "+c" that determines where the zeros are, and that gets differentiated away.
 
  • #10
arildno
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I have always thought that it would be awsome if "x= 1" were the correct answer to every problem but that doesn't work either!

:rofl:
 
  • #11
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My knowledge of calculus tells me that it is because 0 = ax2 + bx + c is not a function, and is only fulfilled by 2 real values, but I want there to be deeper reason. =|

Because when you differentiate first, and then you go back with an integration, you lose the constant term.
That is
[itex]\int {{f}\ '(x)} = F(x) +K[/itex]

so you obtain
[itex]ax^2+bx+K[/itex]
which is not your original expression.
 
  • #12
Hurkyl
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Because when you differentiate first, and then you go back with an integration, you lose the constant term.
That is
[itex]\int {{f}\ '(x)} = F(x) +K[/itex]

so you obtain
[itex]ax^2+bx+K[/itex]
which is not your original expression.
I think you're missing the misconception that led to his question, which is basically:
  1. I have an equation in x
  2. If two things are equal, the derivative of both sides are equal
  3. Therefore, I differentiate my equation to get a new equation that is easier to solve
Simply reinforcing "your conclusion is wrong" doesn't help -- he already knows that. What he needs is to have explained what went wrong in his reasoning. e.g. having some of the subtleties of mathematical grammar explained.
 
  • #13
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Solving F'(x)=0 would find the vertex of the parabola, or where the slope(derivative) is equal to zero. Anytime a continuous curve changes direction, the slope must be equal to zero somewhere. On a cubic equation, F'(x)=0 would net you the max and min of the cubic etc..
 
  • #14
HallsofIvy
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Solving F'(x)=0 would find the vertex of the parabola, or where the slope(derivative) is equal to zero. Anytime a continuous curve changes direction, the slope must be equal to zero somewhere. On a cubic equation, F'(x)=0 would net you the max and min of the cubic etc..
This is not true. The cubic [itex](1/3)x^3+ x^2+ 2x+ 1[/itex] has derivative [itex]x^2+ 2x+ 2[/itex] which is never 0. It has NO max and min.
 
  • #15
Char. Limit
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This is not true. The cubic [itex](1/3)x^3+ x^2+ 2x+ 1[/itex] has derivative [itex]x^2+ 2x+ 2[/itex] which is never 0. It has NO max and min.

However, IF the maximum and minimum of a cubic exist, then his method is the way to find them.
 
  • #16
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I think you're missing the misconception that led to his question, which is basically:
  1. I have an equation in x
  2. If two things are equal, the derivative of both sides are equal
  3. Therefore, I differentiate my equation to get a new equation that is easier to solve
Simply reinforcing "your conclusion is wrong" doesn't help -- he already knows that. What he needs is to have explained what went wrong in his reasoning. e.g. having some of the subtleties of mathematical grammar explained.

My misconception was actually that the zeros of a function are the same as the mins of a function, which is obviously not correct.

ps: I can't beleive people are still responding to this thread...lol -.-
 
  • #17
My misconception was actually that the zeros of a function are the same as the mins of a function, which is obviously not correct.

Sometimes it is true(when the discriminant is 0). You can use the Newton method to approximate the roots of any function though.
 
  • #18
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This is not true. The cubic [itex](1/3)x^3+ x^2+ 2x+ 1[/itex] has derivative [itex]x^2+ 2x+ 2[/itex] which is never 0. It has NO max and min.

HAHAHA, sorry should've been more precise I guess. Was trying to make a point in general. Next time I post I will be sure to submit a proof along with my analysis. LoL
 
  • #19
lavinia
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The solutions are discrete. What about for more than one variable?
 

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