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    Okay, pretty much covered this topic, just the last one I am stuck on;

    20.)
    a.)
    Expand (1+2x)^(-1/2), |x|<(1/2), in ascending powers of x up to and including the term in x³.

    b.)
    Hence, show that for small values of x,

    2-5x/√(1+2x) ≈ 2-7x+8x²-(25/2)x³

    c.)
    Solve the equation;

    2-5x/√(1+2x) = √3.

    d.)
    Use answers to b & c to find an approximate value for √3.

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    2-5x/√(1+2x) = √3
    or
    (2-5x)/√(1+2x) = √3
    ?
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    (Original post by mik1w)
    2-5x/√(1+2x) = √3
    or
    (2-5x)/√(1+2x) = √3
    ?
    (2-5x)/√(1+2x) = √3
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    Any takers?
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    (Original post by Slice'N'Dice)
    (2-5x)/√(1+2x) = √3
    (2-5x)/√(1+2x) = √3

    put y = 1+2x

    (2-5x)/√y = √3
    2/√y - 5x/√y = √3
    2 - 5x = √3√y
    2 - 5x = √(3y)
    2 - 5x = √(3 + 6x)
    25x^2 - 20x + 4 = 3 + 6x
    25x^2 - 26x + 1 = 0
    (25x - 1)(x - 1) = 0
    x = 1 or 1/25

    put both solutions back in to check that neither are "false" from the squaring we did earlier, that should be okay.
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    Alrighty will do. Thankyou m.
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    I haven't learnt the rest. only upto C3 so far
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    (Original post by Slice'N'Dice)
    Okay, pretty much covered this topic, just the last one I am stuck on;

    20.)
    a.)
    Expand (1+2x)^(-1/2), |x|<(1/2), in ascending powers of x up to and including the term in x³.

    b.)
    Hence, show that for small values of x,

    2-5x/√(1+2x) ≈ 2-7x+8x²-(25/2)x³

    c.)
    Solve the equation;

    2-5x/√(1+2x) = √3.

    d.)
    Use answers to b & c to find an approximate value for √3.

    (1+2x)^(-1/2)=1-x+(-1/2)(-3/2)/2!(2x)^2+(-1/2)(-3/2)(-5/2)/3!(2x^3)
    =1-x+3x^2/2-5/2x^3
    for small values
    2-5x/√(1+2x)=(2-5x)(1-x+3x^2/2-5/2x^3)
    =2-2x+3x^2-5x^3-5x+5x^2-15/2x^3
    =2-7x+8x^2-(25/2)x^3..............(1)
    ill take the 1/25 as a correct solution for 2-5x/√(1+2x) = √3.
    so approx for √3 given by
    2-(7/25)+8/625-1/1250 .............using (1) with x=1/25
    =(2500-350+16-1)/1250
    =2165/1250
    =433/250
    =1.732
 
 
 
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