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    Find the first four terms of (1-3x)^{3/2} by substituting a suitable value of x, find an approximation to 97^{3/2}

    I'm pretty sure I've done a question like this before from C2 but i can't remember how to determine a suitable x value. I tried making 1-3x =97 but this was wrong. Any help would be appreciated
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    Try 1-3x=97k for any suitable k

    For example:

    If k=1, 1-3x=97.

    If k=1/10, 1-3x=9.7.

    etc.
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    Remember x has to be between -1/3 and 1/3. Also remember that you can manipulate 0.97^(3/2) to get 97^(3/2) using laws of indicies.
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    (Original post by raiden95)
    Find the first four terms of (1-3x)^{3/2} by substituting a suitable value of x, find an approximation to 97^{3/2}

    I'm pretty sure I've done a question like this before from C2 but i can't remember how to determine a suitable x value. I tried making 1-3x =97 but this was wrong. Any help would be appreciated
    \displaystyle 97^\frac{3}{2} = \left(0.97\times 100\right)^{\frac{3}{2} }= 0.97^{\frac{3}{2}} \times 1000
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    It would be easier to equate the 1-3x to something more smaller. This is so that when you input it in to the binomial expansion it will make it that little bit easier.

    try equating 1-3x to 0.97
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    (Original post by raiden95)
    Find the first four terms of (1-3x)^{3/2} by substituting a suitable value of x, find an approximation to 97^{3/2}

    I'm pretty sure I've done a question like this before from C2 but i can't remember how to determine a suitable x value. I tried making 1-3x =97 but this was wrong. Any help would be appreciated
    I'm sure we had this question a couple of weeks ago!

    remember you need -1<3x<1 for the infinite binomial expansion to be valid, so you'll need to set 1-3x = 97k as suggested by the other posters where k is some suitable value that will give you a convenient multiplier when raised to the power 3/2.
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    (Original post by davros)
    I'm sure we had this question a couple of weeks ago!

    remember you need -1<3x<1 for the infinite binomial expansion to be valid, so you'll need to set 1-3x = 97k as suggested by the other posters where k is some suitable value that will give you a convenient multiplier when raised to the power 3/2.
    I put k =0.1 and got x=-2.9
    Then i tried k=0.01 and got x=0.01
    So its less than 1 now and i know it must be 0.01 but k isn't always the same as x, so i use the x not k right always when substituting?
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    (Original post by aznkid66)
    Try 1-3x=97k for any suitable k

    For example:

    If k=1, 1-3x=97.

    If k=1/10, 1-3x=9.7.

    etc.
    Is my method correct?
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    (Original post by raiden95)
    Is my method correct?
    You need to set 1-3x = 0.97 and adjust using the pointers that notnek has given above
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    (Original post by Indeterminate)
    You need to set 1-3x = 97 and adjust using the pointers that notnek has given above
    Can you look at post 7 please
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    (Original post by raiden95)
    Can you look at post 7 please
    notnek's post number 4 actually gives you a clue what to choose for x
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    (Original post by davros)
    notnek's post number 4 actually gives you a clue what to choose for x
    I thought if its less than 1 thats what i choose?

    And to be honest i really don't understand it
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    (Original post by notnek)
    \displaystyle 97^\frac{3}{2} = \left(0.97\times 100\right)^{\frac{3}{2} }= 0.97^{\frac{3}{2}} \times 1000
    I don't quite understand how this works or how i can use it to find x
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    (Original post by raiden95)
    I don't quite understand how this works or how i can use it to find x
    Can you approximate 0.97 to that power?
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    (Original post by raiden95)
    I thought if its less than 1 thats what i choose?

    And to be honest i really don't understand it
    What value of x would give you 0.97 inside the bracket?

    How does 0.97 relate to 97?

    How does 0.97^{3/2} relate to 97^{3/2}?
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    Wheres this question from??
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    (Original post by raiden95)
    I don't quite understand how this works or how i can use it to find x
    It works because:

    (ab)^n = a^n b^n

    \displaystyle 97^\frac{3}{2} = 100^\frac{3}{2} \times 0.97^\frac{3}{2}
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    (Original post by Mr M)
    It works because:

    (ab)^n = a^n b^n

    \displaystyle 97^\frac{3}{2} = 100^\frac{3}{2} \times 0.97^\frac{3}{2}
    Just to clarify, we chose 0.97 because its less than 1? Or is it always 2dp
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    (Original post by raiden95)
    Just to clarify, we chose 0.97 because its less than 1? Or is it always 2dp
    It gives a value of x that is in the range of values that the expansion is valid for
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    (Original post by raiden95)
    Just to clarify, we chose 0.97 because its less than 1? Or is it always 2dp
    You are trying to make 1-3x equal to something. 0.97 is sensible as this means x = 0.01.

    Remember the binomial expansion of (1+y)^n is only valid for |y| &lt; 1 so you need to choose x carefully.
 
 
 
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