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    Hi guys

    hope you're well.

    I know, that when you have x^4 * x^8 = x^12

    What about when you have: x^4 + x^8 = ??

    I have no idea which formula to use! :/

    Help would certainly be appreciated Reps available No, really
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    You can't really do anything with x^4 + x^8. It's already in its simplest form. Is this part of a question? Or just a general query?
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    (Original post by Daniel Freedman)
    You can't really do anything with x^4 + x^8. It's already in its simplest form. Is this part of a question? Or just a general query?
    Firstly, thank you for actually replying
    I made that up, to save myself from putting up the question I need to actually answer(not one to cheat :rolleyes: )

    okay, let's try another:
    x^1/4 - 3x^1/4 - 10x^-3/4 =0
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    (Original post by dream123)
    Firstly, thank you for actually replying
    I made that up, to save myself from putting up the question I need to actually answer(not one to cheat :rolleyes: )

    okay, let's try another:
    x^1/4 - 3x^1/4 - 10x^-3/4 =0
    you can add numbers with the same index so your equation can be reduced to:

    -2x^1/4=10x^-3/4

    raise both sides to the power of 4 so the fractions in the indices cancel

    (-2x^1/4)^4=(10x^-3/4)^4

    16x=10000x^-3

    then just rearrange to give:

    x=(10000/16)^1/4=5
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    (Original post by TheBigRC)
    you can add numbers with the same index so your equation can be reduced to:

    -2x^1/4=10x^-3/4

    raise both sides to the power of 4 so the fractions in the indices cancel

    (-2x^1/4)^4=(10x^-3/4)^4

    16x=10000x^-3

    then just rearrange to give:

    x=(10000/16)^1/4=5
    I understand it all very nicely upto the part in green , where does ^1/4 come from?
    I really appreciate this (rep your way once I've established the green part )
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    In general (see below), there isn't really a rule for simplifications of the sums of powers.
    But usually there'll be some kind of trick you can use to simplify (and there are lots of weird examples)

    For your question, consider multiplying both sides by x^a, so then it becomes (using the product of powers rule)
    

x^{a + 1/4} - 3 x^{a + 1/4} - 10 x^{a - 3/4} = 0

    Now we can choose a to be whatever we want, and the equation will still hold, right?
    Is there a value (or values) of a that will simplify the equation?

    Hint: Yes.

    PS. A weird non general power examples:

    12^2 + 33^2 = 1233
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    (Original post by dream123)
    I understand it all very nicely upto the part in green , where does ^1/4 come from?
    I really appreciate this (rep your way once I've established the green part )
    ok, so you understand up until:

    16x=10000x^-3

    multiply both sides by x^3 to get:

    16*(x^(1+3))=10000*(x^(3-+3))

    16x^4=10000

    divide both sides by 16

    x^4=10000/16

    then, as before raise both side to 1/4 to clear the x^4 on the LHS

    (x^4)^1/4=(10000/16)^1/4

    therefore (drum roll...)

    x=(10000/16)^1/4=5
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    (Original post by TheBigRC)
    ok, so you understand up until:

    16x=10000x^-3

    multiply both sides by x^3 to get:

    16*(x^(1+3))=10000*(x^(3-+3))

    16x^4=10000

    divide both sides by 16

    x^4=10000/16

    then, as before raise both side to 1/4 to clear the x^4 on the LHS

    (x^4)^1/4=(10000/16)^1/4

    therefore (drum roll...)

    x=(10000/16)^1/4=5
    GOT IT! Oh you little genius!
    Fom the point in red, i decided to root it by 4 (im not sure what name is given to the cube root equivalent of 4 :confused: ).
    And yes, surely, my answer = 5

    Hopefully, I can solve the question

    Rep your way!!
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    (Original post by dream123)
    GOT IT! Oh you little genius!
    Fom the point in red, i decided to root it by 4 (im not sure what name is given to the cube root equivalent of 4 :confused: ).
    And yes, surely, my answer = 5

    Hopefully, I can solve the question

    Rep your way!!
     \sqrt[4]{x} is called the "4th-root" of x and is equivalent to x^{1/4}

    infact, in general:

     \sqrt[n]{x}=x^{1/n} and is called the "nth-root" of x

    good luck with solving your questions!
 
 
 
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