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Math Help - Why not solve the quadratic this way?

  1. #1
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    Why not solve the quadratic this way?

    I have (289^2)+(-1666)+(2401)=(x^2-5)

    then I, (289^2)+((-1666)+(2401))=(x^2-5)

    then I have, (289^2)+(735)=(x^2-5) then subtract (x^2) and add 5 to both sides.

    So now I have (288x^2)+740=0

    So then, subtract 740 from both parts to get (288x^2)=-740

    Finally do this -740/288=x^2, so x=(-2.5)^(1/2) which is imaginary. But the equation has only real solutions. But it looks like the answer was derived using all correct principles...?
    Last edited by mr fantastic; April 19th 2011 at 04:42 AM. Reason: Title.
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  2. #2
    Super Member TheChaz's Avatar
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    Quote Originally Posted by bournouli View Post
    I have (289^2)+(-1666)+(2401)=(x^2-5)

    then I, (289^2)+((-1666)+(2401))=(x^2-5)

    then I have, (289^2)+(735)=(x^2-5) then subtract (x^2) and add 5 to both sides.

    So now I have (288x^2)+740=0

    So then, subtract 740 from both parts to get (288x^2)=-740

    Finally do this -740/288=x^2, so x=(-2.5)^(1/2) which is imaginary. But the equation has only real solutions. But it looks like the answer was derived using all correct principles...?
    Why are there so many PARENTHESIS?!?!
    Is the original equation 289^2 - 1666 + 2401 = x^2 - 5 ???
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  3. #3
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    Prove It's Avatar
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    Maybe because 289^2 doesn't have an x^2 term, so they are not like terms and can not be added/subtracted...
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Maybe because 289^2 doesn't have an x^2 term, so they are not like terms and can not be added/subtracted...
    Sorry it was (289x^2).
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  5. #5
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    Quote Originally Posted by bournouli View Post
    Sorry it was (289x^2).
    If and only if the original equation was:

    289x^2 - 1666x + 2401 = x^2 - 5

    then you'll get the solutions x = 3 or x = 401/144

    ... but my crystal sphere is a little bit dusty
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