Solve the system

$\displaystyle

x^2-4\sqrt{3x-2}+6=y$

$\displaystyle

y^2-4\sqrt{3y-2}+6=x

$

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- Jun 23rd 2008, 08:09 PMmathwizardSolve the system
Solve the system

$\displaystyle

x^2-4\sqrt{3x-2}+6=y$

$\displaystyle

y^2-4\sqrt{3y-2}+6=x

$ - Jun 24th 2008, 12:53 AMOpalg
There is an obvious solution x = y = 2. In fact, this is the only solution.

To see why, let $\displaystyle f(x) = x^2-4\sqrt{3x-2}+6$ (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.