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    given that 2x^2 - x^2/3 can be written in the form 2x^p - x^q, find values for p & q

    Okay, so I split the equation in half and the denominator is the √x

    2x^2 ÷ √x and -x^3/2 ÷ √x

    Now I use the laws of indices. So √x is x^1/2

    I have a feeling this is where I went wrong, is it -1/2?
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    (Original post by samantham999)
    given that 2x^2 - x^2/3 can be written in the form 2x^p - x^q, find values for p & q...
    Right off the bat the question doesn't lead anywhere unless I'm misinterpreting it. It's already in that form; p=2 and q=2/3. Where did the division by root x come from?
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    (Original post by RDKGames)
    Right off the bat the question doesn't lead anywhere unless I'm misinterpreting it. It's already in that form; p=2 and q=2/3. Where did the division by root x come from?
    crap, I split the fraction into two different ones with a common denominator? the question asks write down the value of p & q

    so i did 2x^2 over root x?
    -x^3/2 over root x?
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    (Original post by samantham999)
    crap, I split the fraction into two different ones with a common denominator? the question asks write down the value of p & q

    so i did 2x^2 over root x?
    -x^3/2 over root x?
    What's the full question??

    I'm assuming it is to write \frac{2x^2-x^{-3/2}}{\sqrt{x}} then yeah root of x is x to the power of a half. It cannot be negative since then it would be 1 over root x.
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    (Original post by RDKGames)
    What's the full question??

    I'm assuming it is to write \frac{2x^2-x^{-3/2}}{\sqrt{x}} then yeah root of x is x to the power of a half. It cannot be negative since then it would be 1 over root x.
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    So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

    From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
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    (Original post by samantham999)
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    So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

    From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
    The expression in the question can be written as the following  \displaystyle x^{-\frac{1}{2}}(2x^2-x^{-\frac{3}{2}}) .
    Now you just use laws of indices.
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    (Original post by samantham999)


    So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

    From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
    Yes
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    (Original post by RDKGames)
    Yes

    so for 2x^2 over x^1/2 i got 2x^1/2 ?

    -x^3/2 over 2^1/2 i got 4/2 ? which is 2/1 ?
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    (Original post by samantham999)
    so for 2x^2 over x^1/2 i got 2x^1/2 ?
    No.
    You have 2-\frac{1}{2} as the exponent, surely you can do this GCSE problem subtraction.

    -x^3/2 over 2^1/2 i got 4/2 ? which is 2/1 ?
    Where did the 2^{1/2} come from?? How did you get 4/2????

    -x^{\frac{3}{2}-\frac{1}{2}} when dividing by root x.
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    (Original post by samantham999)
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    So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

    From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
    yes
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    (Original post by RDKGames)
    No.
    You have 2-\frac{1}{2} as the exponent, surely you can do this GCSE problem subtraction.



    Where did the 2^{1/2} come from?? How did you get 4/2????

    -x^{\frac{3}{2}-\frac{1}{2}} when dividing by root x.

    but why isn't it 2x^2 divide x^1/2 ? don't i do 2 - 1/2 like m-n ? i got 2x^3/2 ?
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    (Original post by samantham999)
    but why isn't it 2x^2 divide x^1/2 ? don't i do 2 - 1/2 like m-n ? i got 2x^3/2 ?
    Yeah it's that for the exponent. And 2-1/2 is not equal to 1/2 as you said first
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    Sqrt(x) is the same as x^(1/2)
    Dividing by x^(1/2) is the same as multiplying by x^(-1/2)

    So you end up with (2x^2 - x^(2/3)) * x^(-1/2)

    Then literally all you have to do is expand the brackets, remembering this law of indices:

    x^a * x^b = x^(a+b).

    Just in case you haven't got it yet.
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    (Original post by RDKGames)
    Yeah it's that for the exponent. And 2-1/2 is not equal to 1/2 as you said first
    for -x^3/2 - 1/2 why do i keep getting 4/2 ? which is 2/2 which is 1 ?
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    (Original post by samantham999)
    for -x^3/2 - 1/2 why do i keep getting 4/2 ? which is 2/2 which is 1 ?
    lol

    "4/2 which is 2/2" no it isn't.

    You should be getting 2/2 which is 1 because 3/2-1/2=2/2, dunno where you're pulling the 4 from
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    (Original post by RDKGames)
    lol

    "4/2 which is 2/2" no it isn't.

    You should be getting 2/2 which is 1 because 3/2-1/2=2/2, dunno where you're pulling the 4 from
    wow i really need to take care, so many stupid mistakes. i got 2/2 which is 1 so its -x^1 ?
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    (Original post by samantham999)
    wow i really need to take care, so many stupid mistakes. i got 2/2 which is 1 so its -x^1 ?
    Yes
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    (Original post by samantham999)
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    So i get 2x^2 over x^1/2 and -x^3/2 over x^1/2

    From there I got completely confused. Do i use the indices rule of x^m divide x^n = x^m-n ?
    (2x^2)/(x^1/2)=2x^3/2 because you just take the powers away from each other when yiou divide i.e. 2-1/2=1.5 which is 3/2.
    Then you do the same for the other side:
    (-x^3/2)/(x^1/2)=-x^2/2 2/2=1
    Therefore the ful answer is :
    2x^3/2-x^1
 
 
 
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