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Math Help - Question on method

  1. #1
    Senior Member I-Think's Avatar
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    Question on method

    Determine all real numbers such that (x^2-3x)^2=4-3(3x-x^2)
    Last edited by I-Think; July 6th 2009 at 04:57 PM. Reason: Unreadable
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  2. #2
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    Please clean up that posting. It is unreadable.
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  3. #3
    Senior Member I-Think's Avatar
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    Problem fixed.
    Last edited by I-Think; July 6th 2009 at 07:21 PM.
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  4. #4
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    Quote Originally Posted by I-Think View Post
    Determine all real numbers such that (x^2-3x)^2=4-3(3x-x^2)

    let u = 3x-x^2

    u^2 = 4 - 3u

    u^2 + 3u - 4 = 0

    (u + 4)(u - 1) = 0

    first solution ...

    u = -4

    3x - x^2 = -4

    0 = x^2 - 3x - 4

    0 = (x - 4)(x + 1)

    x = 4 , x = -1

    second solution ...

    u = 1

    3x - x^2 = 1

    0 = x^2 - 3x + 1

    x = \frac{3 \pm \sqrt{5}}{2}
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  5. #5
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    Hello, I-Think!

    That substitution is clever, but it can be solved head-on . . .


    Solve for x\!:\;\; (x^2 -3x)^2 \:=\:-3(3x-x^2)
    The simplified equation is: . x^4 - 6x^3 + 6x^2 + 9x - 4 \;=\;0

    Factor: . (x+1)(x-4)(x^2-3x+1) \;=\;0

    Therefore: . x \;=\;-1,\:4,\:\tfrac{3\pm\sqrt{5}}{2}

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  6. #6
    Senior Member I-Think's Avatar
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    Thanks for replying and managing to read the undecipherable text.
    Here's the problem I'm having with the question.

    My solution:
    (x^2-3x)^2=4-3(3x-x^2)
    (x^2-3x)^2-4=-3(3x-x^2)
    (x^2-3x+2)(x^2-3x-2)=-3(3x-x^2)

    Case 1: x^2-3x+2=-3, x^2-3x-2=(3x-x^2)<br />
    x^2-3x+2=-3\rightarrow{x^2-3x+5=0}
    x= a complex number
     x^2-3x-2=3x-x^2
    x^2-3x-1=0
    x=\frac{3\pm{\sqrt{13}}}{2}

    Case 2: x^2-3x+2=3x-x^2, x^2-3x-2=-3
    x^2-3x+2=3x-x^2
    x^2-3x+1=0
    x=\frac{3\pm{\sqrt{5}}}{2}

    x^2-3x-2=-3
    x^2-3x+1=0
    Same as above

    Solution Set: x=\frac{3\pm{\sqrt{13}}}{2}, \frac{3\pm{\sqrt{5}}}{2}<br />
    Now my problem is, the official solution to this question gives the same answer as skeeter and soroban, but my method gives two different answers.
    But the thing is, they seem to be equally valid, I verified it with a calculator, and I'm wondering if this is possible.
    Can a quartic curve intersect with a quadratic curve 6 times?
    This is the question I would like answered.

    Thanks
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  7. #7
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    the two solutions , \frac{3 \pm \sqrt{13}}{2} , do not work in the original equation ...

    what's that tell you?
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  8. #8
    Senior Member I-Think's Avatar
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    My calculator has betrayed me...
    My apologies for wasting the forum's time due to a calculator error.
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