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    So i had this question
    Show that
    

f(x) = (3-4\sqrt{x})^2

Show that f(x) = 9x^{-1/2} + Ax^{1/2} + B

where A and B are constants

    I done the first bit - expanding
    Which leaves me with
     \frac{9-24\sqrt{x} +16x}{\sqrt{x}} \

    I know the answer etc. as it saids it in the question, I just don't undertstand how you divide by a square root
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    bump - fixed formats
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    Think about what root x would be if you wrote it as x to a power
    Then actually try to divide each term of the numerator by this and see what you get - use power laws
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    (Original post by AdeptDz)
    So i had this question
    Show that
    

f(x) = (3-4\sqrt{x})^2

Show that f(x) = 9x^{-1/2} + Ax^{1/2} + B

where A and B are constants

    I done the first bit - expanding
    Which leaves me with
     \frac{9-24\sqrt{x} +16x}{\sqrt{x}} \

    I know the answer etc. as it saids it in the question, I just don't undertstand how you divide by a square root
    Divide each term by \sqrt{x} and remember that \sqrt{x}=x^{1/2} also \frac{1}{a}=a^{-1}

    Also f(x) cannot be in that form unless it is \displaystyle \frac{f(x)}{\sqrt{x}}
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     \frac{9-24\sqrt{x}+16x} {\sqrt{x}} can be written as:  \frac{9} {\sqrt{x}} + \frac{-24\sqrt{x}} {\sqrt{x}} + \frac{16x}{\sqrt{x}} . See if that helps you.

    If you need some extra help with figuring out A:
    Spoiler:
    Show
     \frac{16x}{\sqrt{x}} = 16\sqrt{x}
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    (Original post by RDKGames)
    Divide each term by \sqrt{x} and remember that \sqrt{x}=x^{1/2} also \frac{1}{a}=a^{-1}
    Yh thanks, i understand what you said. So okay it saids the answer to:
     \frac {9}{\sqrt{x}} = 9x^{-1/2}
    So like it is
     9 / x^{1/2} = or is that basically saying 9/a so would it be  9x^{-1/2}
    If that isn't correct then I still don't get it
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    (Original post by BobBobson)
     \frac{9-24\sqrt{x}+16x} {\sqrt{x}} can be written as:  \frac{9} {\sqrt{x}} + \frac{-24\sqrt{x}} {\sqrt{x}} + \frac{16x}{\sqrt{x}} . See if that helps you.

    If you need some extra help with figuring out A:
    Spoiler:
    Show
     \frac{16x}{\sqrt{x}} = 16\sqrt{x}
    Thanks, (@ spoiler) I should know this but I still can't figure out why that is the case

    Edit: Ohh i think i understand now, it is basically like saying  2 \div \sqrt{2} = \sqrt{2} right?
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    And then its like saying  24 \sqrt{x} \div \sqrt{x} is the same as root x divided by root x which is 1? so it would just be 24
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    (Original post by AdeptDz)
    Yh thanks, i understand what you said. So okay it saids the answer to:
     \frac {9}{\sqrt{x}} = 9x^{-1/2}
    So like it is
     9 / x^{1/2} = or is that basically saying 9/a so would it be  9x^{-1/2}
    If that isn't correct then I still don't get it
    That's correct.
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    (Original post by RDKGames)
    That's correct.
    Okay, thanks.. again

    And thank you to everyone above aswell
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    (Original post by AdeptDz)
    So i had this question
    Show that
    

f(x) = (3-4\sqrt{x})^2

Show that f(x) = 9x^{-1/2} + Ax^{1/2} + B

where A and B are constants

    I done the first bit - expanding
    Which leaves me with
     \frac{9-24\sqrt{x} +16x}{\sqrt{x}} \

    I know the answer etc. as it saids it in the question, I just don't undertstand how you divide by a square root
    You've got the answer but in the wrong format, remember root(x) = x^1/2
 
 
 
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