What other ways can you split an equation into two?

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The discussion centers on the concept of "splitting an equation into two," which is deemed not generally meaningful in mathematics. Two examples are provided, but the second example is critiqued for being incorrect, as it does not account for all possible solutions. The first example correctly illustrates that if the product of two numbers is zero, at least one must be zero. The conversation emphasizes that the technique is more about finding zeros of an equation rather than splitting it. Ultimately, the participants agree that clarity in the purpose of the operation is essential.
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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?
 
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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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