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    Hello everyone,

    I'm stuck on this question - when I substitute 0.001 into the expanded form, I get 0.997991.... , which is ten times smaller than the actual answer.

    I'm just wondering where I went wrong - did I do the binomial expansion wrong?
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    (Original post by Electrogeek)
    Hello everyone,

    I'm stuck on this question - when I substitute 0.001 into the expanded form, I get 0.997991.... , which is ten times smaller than the actual answer.

    I'm just wondering where I went wrong - did I do the binomial expansion wrong?
    You haven't done anything wrong.

    The expansion does not directly give you the fifth root you require. Can you see why?
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    (Original post by Electrogeek)
    Hello everyone,

    I'm stuck on this question - when I substitute 0.001 into the expanded form, I get 0.997991.... , which is ten times smaller than the actual answer.

    I'm just wondering where I went wrong - did I do the binomial expansion wrong?
    Firstly note this:
    Sub  x = 0.001 , this gives us \sqrt[5]{1-0.01} = \sqrt[5]{0.99} .

    Then,
    check  \sqrt[5]{99000} = \sqrt[5]{0.99 \times 100000} = \sqrt[5]{0.99} \times \sqrt[5]{100000}

    This gives us:  10 \sqrt[5]{0.99}

    So the answer you are after is indeed 10 times  \sqrt[5]{0.99}
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    (Original post by rashid.mubasher)
    Firstly note this:
    Sub  x = 0.001 , this gives us \sqrt[5]{1-0.01} = \sqrt[5]{0.99} .

    Then,
    check  \sqrt[5]{99000} = \sqrt[5]{0.99 \times 100000} = \sqrt[5]{0.99} \times \sqrt[5]{100000}

    This gives us:  10 \sqrt[5]{0.99}

    So the answer you are after is indeed 10 times  \sqrt[5]{0.99}
    I understand now! I was getting worried then.

    Thanks for the help. 🙂
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    To save me opening another thread, can I have some help with 7b - I don't know how to find which values x is valid for in the new expression.

    Also, I don't know if the combined expression is correct either?
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    In part (b) it looks as though you have copied the first bracketed expression wrongly, with 3 rather than 27 in the denominator of two of the terms. When it comes to multiplying out the two brackets, the approach I would recommend is:

    * On one row, multiply the first expression through by the first term of the second expression.

    * Start a new row. Multiply the first expression through by the second term of the second expression, lining up column-wise with the first row so the x^1 terms form a column and the x^2 terms form a column. You can stop at the x^2 term.

    * Start a third row and do likewise for the third term in the second expression.

    * Now you can add up the columns to find the coefficients of x^0, x^1 and x^2.

    I find this approach cuts down errors (and it will help the examiner to see what you're doing).
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    (Original post by old_engineer)
    In part (b) it looks as though you have copied the first bracketed expression wrongly, with 3 rather than 27 in the denominator of two of the terms. When it comes to multiplying out the two brackets, the approach I would recommend is:

    * On one row, multiply the first expression through by the first term of the second expression.

    * Start a new row. Multiply the first expression through by the second term of the second expression, lining up column-wise with the first row so the x^1 terms form a column and the x^2 terms form a column. You can stop at the x^2 term.

    * Start a third row and do likewise for the third term in the second expression.

    * Now you can add up the columns to find the coefficients of x^0, x^1 and x^2.

    I find this approach cuts down errors (and it will help the examiner to see what you're doing).
    Thanks for the tip and spotting the mistake - I still got the same expression so I guess I just wrote it down wrong.

    So how do you find where x is valid for this expression?
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    (Original post by Electrogeek)
    So how do you find where x is valid for this expression?
    You have two intervals of validity for x. The smallest one wins.
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    (Original post by Mr M)
    You have two intervals of validity for x. The smallest one wins.
    so would I be right in saying |x| < 1.5?
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    The overall validity is set by whichever is the more restrictive of the individual validities.
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    (Original post by Electrogeek)
    so would I be right in saying |x| < 1.5?
    Yes.
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    (Original post by old_engineer)
    The overall validity is set by whichever is the more restrictive of the individual validities.
    Cool - thanks for all the help.
 
 
 
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