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Math Help - Equation with radical

  1. #1
    Newbie Miasmagasma's Avatar
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    Equation with radical

    here's another i can't solve.

    t - 5√t - 14 = 0

    i added 14 to both sides and then squared them. then i minused 196 and am left with t^2 + 15t -196 = 0

    this doesn't work, what have i done wrong?
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  2. #2
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    Quote Originally Posted by Miasmagasma View Post
    here's another i can't solve.

    t - 5√t - 14 = 0

    i added 14 to both sides and then squared them. then i minused 196 and am left with t^2 + 15t -196 = 0

    this doesn't work, what have i done wrong?
    t-14 = 5\sqrt{t}

    t^2 - 28t + 196 = 25t

    t^2 - 53t + 196 = 0
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  3. #3
    Newbie Miasmagasma's Avatar
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    Quote Originally Posted by skeeter View Post
    t-14 = 5\sqrt{t}

    t^2 - 28t + 196 = 25t

    t^2 - 53t + 196 = 0
    cheers but why did my method not work?
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  4. #4
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    Quote Originally Posted by Miasmagasma View Post
    cheers but why did my method not work?
    Because (t - 5 \sqrt{t})^2 \neq t^2 + 15t. If you post all your working on how you got that (wrong) result your error(s) can be explained.
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  5. #5
    Newbie Miasmagasma's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Because (t - 5 \sqrt{t})^2 \neq t^2 + 15t. If you post all your working on how you got that (wrong) result your error(s) can be explained.
    well i thought t times 5√t = 5t which is no doubt wrong. how do you work that out?
    Last edited by mr fantastic; December 9th 2009 at 07:30 PM. Reason: Moved new question to new thread
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  6. #6
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    Quote Originally Posted by Miasmagasma View Post
    here's another i can't solve.

    t - 5√t - 14 = 0

    i added 14 to both sides and then squared them. then i minused 196 and am left with t^2 + 15t -196 = 0

    this doesn't work, what have i done wrong?
    Notice it looks very much like a quadratic.

    So use a dummy variable x^2 = t.

    This would mean x = \sqrt{t}.


    Your equation becomes

    x^2 - 5x - 14 = 0

    (x - 7)(x + 2) = 0

    x = -2 or x = 7.


    Note that, since \sqrt{t} = x, the solution x = -2 is unusable.


    This means, since t = x^2, that t = 49.
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  7. #7
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    Quote Originally Posted by Miasmagasma View Post
    well i thought t times 5√t = 5t which is no doubt wrong. how do you work that out?
    How can t \cdot 5 \sqrt{t} = 5t? If t = 4 do you get a correct result from this? And what then do you think \sqrt{t} \cdot 5 \sqrt{t} is equal to?
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  8. #8
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    As you surmised, t * 5\sqrt{t} \neq 5t. On the other hand, \sqrt{t} * 5\sqrt{t} = 5t.
    Last edited by NOX Andrew; December 10th 2009 at 05:21 PM.
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