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Math Help - Finding a value of an expression

  1. #1
    Junior Member lanierms's Avatar
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    Finding a value of an expression

    x, y, z are all different real numbers.

    When they satisfy the equation: x^2-kyz=y^2-kzx=z^2-kxy,

    find the value of x+y+z+k.

    I already found the value of k though; it's -2.
    But I can't seem to go any further from that.
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  2. #2
    Junior Member lanierms's Avatar
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    (Why isn't anyone replying? Please help!!)
    Here's how I got the value of K:

    x^2-kyz=y^2-kzx

    x^2-y^2=kyz-kzx

    (x+y)(x-y)=-kz(x-y)

    Since x\not=y,

    x+y=-kz.

    If we do the same thing for all three pairs, we get:

    x+y=-kz, y+z=-kx, z+x=-ky.

    If we add all of them together, we get:

    2(x+y+z)=-k(x+y+z)

    k=-2.
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  3. #3
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    x^2 + 2yz = y^2 + 2xz

    \implies x^2 - 2xz = y^2 - 2yz

    \implies x^2 - 2xz + z^2 = y^2 - 2yz + z^2

    \implies (x - z)^2 = (y - z)^2

    \implies x - z = y - z or x - z = -(y - z)

    \implies x = y or x = 2z - y

    Does this help any?
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  4. #4
    Junior Member lanierms's Avatar
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    I really have no idea how I can use those to get the answer.

    x\not=y, because they are all different real numbers.

    Also, if I try the second one, I get:

    x=2z-y

    y^2+2zx=z^2+2xy

    y^2+2z(2z-y)=z^2+2y(2z-y)

    \longrightarrow y^2+4z^2-2yz=z^2+4yz-2y^2

    \longrightarrow 3y^2-6yz+3z^2=0

    \longrightarrow y^2-2yz+z^2=0

    \longrightarrow (y-z)^2=0

    \therefore y=z

    But that shouldn't be correct because as indicated earlier: x\not=y\not=z
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