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Math Help - Solve the system

  1. June 23rd 2008, 08:09 PM #1
    mathwizard
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    Solve the system

    Solve the system

    <br /> x^2-4\sqrt{3x-2}+6=y
    <br /> y^2-4\sqrt{3y-2}+6=x<br />
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  2. June 24th 2008, 12:53 AM #2
    Opalg
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    Quote Originally Posted by mathwizard View Post
    Solve the system

    <br /> x^2-4\sqrt{3x-2}+6=y
    <br /> y^2-4\sqrt{3y-2}+6=x<br />
    There is an obvious solution x = y = 2. In fact, this is the only solution.

    To see why, let f(x) = x^2-4\sqrt{3x-2}+6 (defined for x ≥ 2/3). The function f(x) – x has a minimum value of 0 at x=2 (calculus exercise). Hence f(x) > x whenever x ≠ 2. So if x ≠ 2 and y = f(x) then y > x, and f(y) ≥ y > x. Thus for x ≠ 2 it can never happen that f(x) = y and f(y) = x.
    Last edited by Opalg; June 24th 2008 at 11:16 AM.
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