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    (Original post by RDKGames)
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    Hi

    if a graph doesn’t have roots, then why have imaginary roots? What do they represent if they dont represent the graph crossing the x axis?
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    (Original post by Maths&physics)
    Hi

    if a graph doesn’t have roots, then why have imaginary roots? What do they represent if they dont represent the graph crossing the x axis?
    Because they are still valid solutions to whatever equation you have, though not purely real.

    Never really looked into those solutions graphically, but this video gives some visual interpretation of them.

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    (Original post by RDKGames)
    A graph always has roots; complex, or purely real which is the main focus of this sort of maths up to A-Level or so.
    Unless you are equating graph with "polynomial", this doesn't make a lot of sense. Even then, technically "1" is a polynomial and has no roots.

    [And yes, the OP probably is equating the 2, but still...]
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    (Original post by DFranklin)
    Unless you are equating graph with "polynomial", this doesn't make a lot of sense. Even then, technically "1" is a polynomial and has no roots.

    [And yes, the OP probably is equating the 2, but still...]
    Yeah you're right, should've clarified. I guess they just mean any elementary function with a single variable of degree 1 or more that has an =0 at the end.
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    (Original post by RDKGames)
    Yeah you're right, should've clarified. I guess they just mean any elementary function with a single variable of degree 1 or more that has an =0 at the end.
    Note 1/z = 0 has no real or complex roots.
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    (Original post by DFranklin)
    Note 1/z = 0 has no real or complex roots.
    Having trouble summing up all the functions he means then by the looks of it! Oh well.
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    (Original post by RDKGames)
    Because they are still valid solutions to whatever equation you have, though not purely real.

    Never really looked into those solutions graphically, but this video gives some visual interpretation of them.

    Fantastic! Thank you!
 
 
 
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