What other ways can you split an equation into two?

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Discussion Overview

The discussion revolves around the concept of "splitting an equation into two" and whether this is a meaningful mathematical operation. Participants explore different examples and question the validity of the technique, particularly in the context of identities and finding zeros of equations.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants present examples of splitting equations, such as $$(x-{ x }_{ 1 })({ x }-{ x }_{ 2 })=0$$ and $${ e }^{ x }({ c }_{ 1 }-3{ c }_{ 2 })+{ e }^{ -32x }({ c }_{ 5 }-{ c }_{ 4 })=0$$.
  • Others argue that "splitting an equation in two" is not a generally meaningful operation and question the correctness of the second example provided.
  • One participant suggests that the technique may be valid if the equation is an identity, prompting further inquiry into the implications of this condition.
  • Another participant emphasizes that the process described is more about finding zeros of the original equation rather than splitting it, urging clarity on the goals of the operation.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity and usefulness of the technique of splitting equations. There is no consensus on whether this method yields meaningful results in general.

Contextual Notes

Some participants note that the technique may be valid under specific conditions, such as when dealing with identities or when looking for solutions that hold for all values of ##x##. However, these conditions remain unresolved and are subject to interpretation.

ainster31
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Here are two ways:

$$(x-{ x }_{ 1 })({ x }-{ x }_{ 2 })=0\\ x-{ x }_{ 1 }=0\quad \quad \quad \quad x-{ x }_{ 2 }=0\\ \\$$$$ \\ { e }^{ x }({ c }_{ 1 }-3{ c }_{ 2 })+{ e }^{ -32x }({ c }_{ 5 }-{ c }_{ 4 })=0\\ { c }_{ 1 }-3{ c }_{ 2 }=0\quad \quad \quad \quad \quad { c }_{ 5 }-{ c }_{ 4 }=0$$

Any other ways?
 
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"Splitting an equation in two" is not in general a meaningful mathematical operation. Your second example is incorrect: can you see why? (clue: a = b = 0 is not the only solution to a + b = 0).
 
MrAnchovy said:
"Splitting an equation in two" is not in general a meaningful mathematical operation.

Why not?

MrAnchovy said:
(clue: a = b = 0 is not the only solution to a + b = 0).

Hmmm... you're right. It's weird that my textbook teaches this technique.

What if it's an identity?
 
Last edited:
ainster31 said:
Here are two ways:

$$(x-{ x }_{ 1 })({ x }-{ x }_{ 2 })=0\\ x-{ x }_{ 1 }=0\quad \quad \quad \quad x-{ x }_{ 2 }=0\\ \\$$$$ \\ { e }^{ x }({ c }_{ 1 }-3{ c }_{ 2 })+{ e }^{ -32x }({ c }_{ 5 }-{ c }_{ 4 })=0\\ { c }_{ 1 }-3{ c }_{ 2 }=0\quad \quad \quad \quad \quad { c }_{ 5 }-{ c }_{ 4 }=0$$

Any other ways?

The first situation relies on the fact that if the product of two real numbers is zero, then one of those numbers must be zero.

The second situation is a statement about the linear independence of ##e^x## and ##e^{-32x}##; if ##ae^x+be^{-32x}=0## for all ##x## (i.e. ##ae^x+be^{-32x}## is the zero function), then ##a=b=0##. It's similar to the statement that if ##a_0+a_1 x+...+a_n x^n=0## for all ##x##, then ##a_0=a_1=...=a_n=0##.
 
ainster31 said:
Why not?
Because it does not in general yield anything useful.

ainster31 said:
Hmmm... you're right. It's weird that my textbook teaches this technique.
Perhaps the textbook is looking for solutions which are valid for all values of ## x ##?

ainster31 said:
What if it's an identity?
I don't understand what you mean.
 
Really this is not splitting an equation into two.

It is a way to find zeros of the original equation. Always keep in mind your goals, what are you trying to do. You are not splitting and equation you are looking for zeros.
 
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