Respostas

2014-05-14T12:46:38-03:00
A) x^2 - 6x +5 =0
Δ = (- 6)² - 4 . 1 . 5
Δ = 36 - 20
Δ = 16
X =  \frac{ -(-6)+-\sqrt{16} }{2.1} =>X =  \frac{6+-4}{2}

x' =  \frac{6+4}{2} =  \frac{10}{2} = 5                      x" =  \frac{6-4}{2} =  \frac{2}{2} =1


b) 3x^2-7x+2 = 0
Δ = (- 7)² - 4 . 3 . 2
Δ = 49 - 24
Δ = 25
X =  \frac{-(-7)+- \sqrt{25} }{2.3} =>  X = \frac{7+-5}{6}

x' =  \frac{7+5}{6} =  \frac{12}{6} = 2                   x" =  \frac{7 - 5}{6} =  \frac{2}{6} ou (÷2) =  \frac{1}{3}


c) x^2+4x-5=0
Δ = 4² - 4 . 1 . (- 5)
Δ = 16 + 20
Δ = 36
X =  \frac{-4+-\sqrt{36} }{2.1} =>  \frac{-4+-6}{2}

x' =  \frac{-4+6}{2} =  \frac{2}{2} = 1                  x" =  \frac{-4-6}{2} =  \frac{- 10}{2} = -5

Qualquer dúvida, comente ;)
1 4 1
obrigado
entendi tudinho *-* brigado linda <3
que bom :D , de nada ^^^
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