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A Real Number problem (Posted on 2006-08-16) Difficulty: 2 of 5
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)

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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
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