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Math Help - Quartic Equation with reals

  1. #1
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    Quartic Equation with reals

    If x and y are reals, and x+y=1 and (x^2+y^2)(x^3+y^3)=12, then what is the value of x^2 + y^2 ?
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  2. #2
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    Quote Originally Posted by warriors837 View Post
    If x and y are reals, and x+y=1 and (x^2+y^2)(x^3+y^3)=12, then what is the value of x^2 + y^2 ?
    x^3+y^3=(x+y)(x^2-xy+y^2)=x^2-xy+y^2

    (x^2+y^2)(x^3+y^3)=(x^2+y^2)(x^2-xy+y^2)=12


    Notice that if we square x+y=1 on both sides we get x^2+y^2+2xy=1 Now we have 1+2xy=1, 2xy=0. Therefore xy=0. This gives (x^2+y^2)(x^2-xy+y^2)=(x^2+y^2)^2=12. So (x^2+y^2)=\sqrt{12}
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  3. #3
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    No, I got 3.
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  4. #4
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    Since, x+y=1, then we can replaced y with 1-x. Thus, the result or x^2 + y^2 will be 3.

    See attachment....

    Do you agree?
    Attached Thumbnails Attached Thumbnails Quartic Equation with reals-picture-1.png   Quartic Equation with reals-picture-2.png  
    Last edited by mr fantastic; October 1st 2009 at 01:30 AM. Reason: Merged posts
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  5. #5
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    Given: "x+y=1 and (x^2+y^2)(x^3+y^3)=12, then what is the value of x^2 + y^2 ? "

    Solution: since (x + y) = 1, we can generate identities from it;

    (x + y)^2 = x^2 + 2xy + y^2 = 1

    x^2 + y^2 = 1 - 2xy - - - (1)

    (x+y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3 = 1

    x^3 + y^3 +3xy(x + y) = 1

    x^3 + y^3 + 3xy(1) = 1

    x^3 + y^3 = 1 - 3xy - - - (2)

    From the given expression: (x^2 + y^2)(x^3 + y^3) = 12, we equate (1) and (2)

    we have (1 - 2xy)(1 - 3xy) = 12, expanding by FOIL Method

    1 - 5xy + 6(xy)^2 = 12, rearranging

    6(xy)^2 - 5xy - 11 = 0, by quadratic formula,

    xy = [5 +/- (25 - 4(6)9(-11))^0.5]/2(6)

    roots for xy: xy = (5 +/- 17)/12

    (+) xy = 22/12 or

    (-) xy = -12/12

    for convenience SAY, xy = -1

    then from (1)

    x^2 + y^2 = 1 - 2xy = 1 - 2(-1) = 3.

    ----------------------------------------------

    -
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  6. #6
    Senior Member pacman's Avatar
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    warriors837, my solution confirms your answer,

    your quartic equation 6x^4 - 12x^3 + 11x^2 - 5x + 1 = 12 has indeed two REAL roots, x1 = 1/2 (1 - sqrt(5)) and x2 = 1/2 (1 + sqrt(5))

    graph below was generated, see
    Attached Thumbnails Attached Thumbnails Quartic Equation with reals-ftft.gif  
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  7. #7
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    Could you explain your approach more in depth?
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  8. #8
    Senior Member pacman's Avatar
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    warriors837, that's the most elaborate solution i made in this forum. I can't make it simplier . . .. after all we have the same answer.

    my answer hinges on the the value of xy, please reread my solution, it utilizes a binomial square and cube.
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