Given that x is a real number, determine all possible solutions of:
√(x-1) + √(3-x) = x² - 4x + 6

We know 1 <= x <= 3 
Squaring both sides gives you 2(1 + ¡î(x-1)(3-x))=(x©÷ - 4x + 6)©÷

We know x©÷ - 4x + 6 ranges from 2 to +¡Ä, so that squared ranges from 4 to +¡Ä

We also know that ¡î(x-1)(3-x) ranges from 0 to 1 (maxing out at x=2), so 2(1 + ¡î(x-1)(3-x)) ranges from 2 to 4 (maxing out at x=2)

If the left half ranges from 2 to 4 and the right half from 4 to +¡Ä, the only possible intersection is when both halves equal 4, (making both halves of the original equation equal 2).

Set one side of the original equation to 2 and you get the only solution, x=2.

Posted by Peter on 2006-12-11 20:56:52