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Zeeng
@Zeeng
December 2019
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Resolva:
(x+3)² = (x-1)² + (x-1)²
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Quadrados perfeitos ...
(x+3)² = (x-1)² + (x-1)²
x² + 6x + 9 = x² - 2x + 1 + x² - 2x + 1
x² + 6x + 9 = 2x² - 4x + 2
x² - 2x² + 6x + 4x + 9 - 2 = 0
-x² + 10x + 7 = 0
Δ = 100 - 28
Δ = 72
x = -10 +-√72/-2
x = -10 +-√36.2/-2
x = -10 +-6√2/-2
x' = 5 - 3√2
x'' = 5 + 3√2 S = { 5 - 3√2 , 5 + 3√2 }
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Verified answer
Quadrados perfeitos ...(x+3)² = (x-1)² + (x-1)²
x² + 6x + 9 = x² - 2x + 1 + x² - 2x + 1
x² + 6x + 9 = 2x² - 4x + 2
x² - 2x² + 6x + 4x + 9 - 2 = 0
-x² + 10x + 7 = 0
Δ = 100 - 28
Δ = 72
x = -10 +-√72/-2
x = -10 +-√36.2/-2
x = -10 +-6√2/-2
x' = 5 - 3√2
x'' = 5 + 3√2 S = { 5 - 3√2 , 5 + 3√2 } ok