Zero vs. Root: What Are the Key Differences?

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The discussion clarifies that while "zero" and "root" can often be used interchangeably in the context of functions and equations, there are subtle distinctions. A function's zero refers to the points where the function equals zero, while a root pertains to the solutions of an equation. The term "root" can also apply in broader mathematical contexts, such as square roots, which adds to the potential for confusion. The distinction is noted as important, particularly in algebra, where a polynomial may have multiple roots but only one x-intercept. Overall, understanding these differences is crucial for accurately interpreting mathematical concepts.
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What's the difference between a zero and a root?
cheers.
 
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For a function, there is none. Basically talking about the roots of a function is a fancy way of speaking about the set of points in the domain where the function takes on the value zero. Though, in other contexts, the word root can make sense whereas zero doesn't (e.g. square root, root system, etc.) -- you'll recognize them when you come across them.
 
Technically, an equation has a root while a function has a zero (a "zero" of the function f is a "root" of the equation f(x)= 0). Compuchip is correct that the distinction is not maintained very much but I think it is a shame. The "root" of an equation does not always mean the right side of the equation is "0" and that is the impression that using "root" to mean "zero" of a function gives!
 
Thanks. By root I mean (x-3)(x+4)=0 =>x=3,-4. Not square roots or anything.

I just remembered something being said like (x-2)^3 has three roots but only 1 x-intercept, and then another question which I can't find seemed to imply it was the same case with zeroes and roots.
cheers,
 
Just had me algebra exam and that was a question!

I pretty much put what HallsofIvy said. I asked the lecturer after and he said that zeroes were to do with the function and roots were an algebraic property.

cheers
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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