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    C1 p34 exercise 2B

    4) Solve the following equations, giving each answer in the form k sqrt 2

    (a) 2ysqrt2 - 3 = (5y / sqrt 2) +1

    can somebody show me how you've worked this to get

    y = -4 sqrt 2

    I can do surds and stuff so far but this question isn't going down very well!!

    Thanks,
    Sean
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    Times the whole thing by root(2), and remember that root(2) * root(2) = 2. So on the left it would just become 4y.
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    2y sqrt2 - 3 - 1 = (5y/ sqrt 2)
    2y sqrt2 - 4 = (5y/ sqrt 2)
    2y x (sqrt2) ^ 2 - 4 sqrt 2 = 5y <<<<<<<<<<<<<<<<(multiply by sqrt 2 on LHS)
    4y - 5y = 4 sqrt 2
    -y = 4 sqrt 2
    y = - 4 sqrt 2

    SIMPLE!
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    (Original post by Rup)
    2y sqrt2 - 3 - 1 = (5y/ sqrt 2)
    2y sqrt2 - 4 = (5y/ sqrt 2)
    2y x (sqrt2) ^ 2 - 4 sqrt 2 = 5y <<<<<<<<<<<<<<<<(multiply by sqrt 2 on LHS)
    4y - 5y = 4 sqrt 2
    -y = 4 sqrt 2
    y = - 4 sqrt 2

    SIMPLE!
    Thanks :] makes sense now

    i'll post up 1 last 1 which i can't get my head around can u have a look pls :]
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    (Original post by Bleak Lemming)
    Thanks :] makes sense now

    i'll post up 1 last 1 which i can't get my head around can u have a look pls :]
    oks
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    (Original post by Rup)
    oks
    Ok I've got as far as

    z (2\sqrt{2}) = 12



z (2\sqrt{2}) = 12



2z\sqrt{2} = 12



z\sqrt{2} = 6

    This isn't going the right way here :s



    EDIT: Wrong post!!
 
 
 
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