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Math Help - Solve the equation system

  1. #1
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    Solve the equation system

    Solve the equation system:
    x+y=1

    x^5+y^5=31
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  2. #2
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    Krizalid's Avatar
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    Quote Originally Posted by james_bond View Post
    Solve the equation system:
    x+y=1

    x^5+y^5=31
    Power to 5 the first equation:

    (x + y)^5 = x^5 + 5x^4 y + 10x^3 y^2 + 10x^2 y^3 + 5xy^4 + y^5.

    So \left( {x^5 + y^5 } \right) + \left( {5x^4 y + 5xy^4 } \right) + \left( {10x^3 y^2 + 10x^2 y^3 } \right).

    And 31 + 5xy\left( {x^3 + y^3 } \right) + 10x^2y^2\left( {x+y } \right).

    May be you can go on from there.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by james_bond View Post
    Solve the equation system:
    x+y=1

    x^5+y^5=31
    I'd recommend the brute force approach.

    y = 1 - x

    x^5 + (1 - x)^5 = 31

    x^5 + 1 - 5x + 10x^2 -10x^3 + 5x^4 - x^5 = 31

    5x^4 - 10x^3 + 10x^2 - 5x - 30 = 0

    x^4 - 2x^3 + 2x^2 - x - 6 = 0

    The rational root theorem gives -1 and 2 as zeros. So do long or synthetic division, which will give you a quadratic to solve at the end.

    -Dan
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  4. #4
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    Step 1 (The Krizalid Operation)
    (x+y)^5 = 1
    x^5+y^5 + 5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5 = 1
    (x^5+y^5)+(5x^4y+5xy^4)+(10x^3y^3+10x^2y^3)=1
    31 + 5xy(x^3+y^3)+10x^2y^2(x+y)=1
    31+5xy(x+y)(x^2-xy+y^2)+10x^2y^2=1
    5xy(x^2-xy+y^2)+10x^2y^2=-30
    5x^3y+5x^2y^2+5xy^3=-30
    x^3y+x^2y^2+xy^3 = -6 . . . [1]

    Step 2
    x^5+y^5= 31
    (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)=31
    (x^4+y^4) - (x^3y+xy^3 - x^2y^2) = 31 . . . (use [1])
    (x^4+y^4) - (-6 - x^2y^2 - x^2y^2) = 31
    x^4+2x^2y^2+y^4 = 25
    (x^2+y^2)^2=25
    x^2+y^2 = \pm 5 . . . [2]

    Step 3
    x+y=1
    x^2+2xy+y^2 = 1 . . . (use [2])
    \pm 5 + 2xy = 1
    xy = 3,2 . . . [3]

    Step 4
    x-y = ?
    x^2 - 2xy +y^2 = ?^2 . . . (use [3]) . . . (use 2)
    ?^2 = \pm 5 - 3,2 = 2,3,-7,-8

    So you just have to solve,
    x+y=1
    x-y=c
    Where c is one of those possible values.
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