Respostas

2014-04-16T18:38:54-03:00
F(x) = x² - 3x + 2
f(x) = 0

f(x) = x² - 3x + 2 = 0

Δ = b² - 4.a.c
Δ = (-3)² - 4.1.2
Δ = 9 - 8
Δ = 1

x = - b ± √Δ / 2a
x = - (-3) ± √1 / 2.1
x = 3 ± 1 / 2
x¹ = 3 + 1 / 2 = 4 / 2 = 2
x² = 3 - 1 / 2 = 2 / 2 = 1

x = 1, 2


2014-04-16T18:40:02-03:00
Basta Aplicar Bháskara.
 \frac{-b(+-) \sqrt{ b^{2}-4ac } }{2a}  \\  \\  \frac{-(-3)(+-) \sqrt{ -3^{2}-4*1*2 } }{2*1}  \\  \\  \frac{3(+-) \sqrt{ 9-8 } }{2} =  \frac{3(+-)1}{2}    \\  \\  x_{1}= \frac{3+1}{2} = \frac{4}{2}=2  \\  \\   x_{2}= \frac{3-1}{2}= \frac{2}{2}=1

Alternativa C
=)
1 5 1