 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  A Real Number problem (Posted on 2006-08-16) Given that x is a real number, determine all possible solutions of:
√(x-1) + √(3-x) = x² - 4x + 6

 See The Solution Submitted by K Sengupta Rating: 4.0000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) It's easier than you make it | Comment 8 of 9 | We know 1 <= x <= 3
Squaring both sides gives you 2(1 + ��(x-1)(3-x))=(x�� - 4x + 6)��

We know x�� - 4x + 6 ranges from 2 to +��, so that squared ranges from 4 to +��

We also know that ��(x-1)(3-x) ranges from 0 to 1 (maxing out at x=2), so 2(1 + ��(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 +��, 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 Please log in:

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