Thread: Solve the system

Solve the system

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

Originally Posted by mathwizard

Solve the system

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

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.