Respostas

2014-03-24T13:35:56-03:00
√(16-x²) = √(12-(x-1)²) + √3

√(16-x²) = √(12-(x²-2x+1)) + √3

√(16-x²) = √(12-x²+2x-1) + √3

(√(16-x²))² = (√(12-x²+2x-1) + √3)²

16-x² = 12-x²+2x - 1 + (2√3 .(12-x²+2x-1)) + 3

16 = 12 + 2x -1 + (√12 . (12-x²+2x-1)) + 3

16 = 12 + 2x -1 + 12√12 - x²√12 + x2√12 -√12 + 3

16 -12 +1 -3 =  2x + 12√12 - x²√12 + x2√12 -√12 

2 = 2x + 12√12 - x²√12 + x2√12 -√12 

2x + 12√12 - x²√12 + x2√12 -√12 -2 = 0

√12x² + x.(2√12+2) + 11√12 -2 = 0

bhaskara:

(2√12+2)² - 4 . √12 . (11√12 -2) = Δ

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