calcule caso exista em r , a raiz da função. x²+6x=0
Soluções para a tarefa
Resposta:
a)
y = x² - 1
0 = x² - 1
1 = x²
x² = 1
x = √1
x = + 1 e x = - 1
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b)
y= x² + 3x + 2
0 = x² + 3x + 2
x² + 3x + 2 = 0
a = 1; b = 3; c = 2
Δ = b² - 4ac
Δ = 3² - 4.1.2
Δ = 9 - 8
Δ = 1
x = - b +/- √Δ = - 3 + / - √1
2a 2.1
x = - 3 + 1 = - 2/2 = - 1
2
x = - 3 - 1 = - 4/2 = - 2
2
R.: x = - 1 e x = - 2
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c)
y= x² + x - 2
0 = x² + x - 2
x² + x - 2 = 0
a = 1; b = 1; c = - 2
Δ = b² - 4ac
Δ = 1² - 4.1.( -2)
Δ = 1 + 8
Δ = 9
x = - b +/- √Δ = - 1 +/- √9
2a 2.1
x = - 1 + 3 = 2/2 = 1
2
x = - 1 - 3 = - 4/2 = - 2
2
R.: x = 1 e x = - 2
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d)
y= x² - 6x + 9
0 = x² - 6x + 9
x² - 6x + 9 = 0
a = 1; b = - 6; c = 9
Δ = b² - 4ac
Δ = (-6)² - 4.1.9
Δ = 36 - 36
Δ = 0
x = - b + / - √Δ = - ( - 6) +/ - 0
2a 2.1
x = 6 = 3
2
R.: x = 3
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e)
y= x² - 4x + 3
0 = x² - 4x + 3
x² - 4x + 3 = 0
a = 1; b = - 4; c = 3
Δ = b² - 4ac
Δ = (-4)² - 4.1.3
Δ = 16 - 12
Δ = 4
x = - b +/- √Δ = - ( - 4) +/- √4
2a 2.1
x = 4 + 2 = 6/2 = 3
2
x = 4 - 2 = 2/2 = 1
2
R.: x = 1 e x = 3
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f)
y= x² + 4x + 3
0 = x² + 4x + 3
x² + 4x + 3 = 0
a = 1; b = 4; c = 3
Δ = b² - 4ac
Δ = 4² - 4.1.3
Δ = 16 - 12
Δ = 4
x = - b +/- √Δ = - 4 + / - √4
2a 2.1
x = - 4 + 2 = - 2/2 = - 1
2
x = - 4 - 2 = - 6/2 = - 3
2
R.: x = - 1 e x = - 3
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g)
y= x² - x - 2
x² - x - 2 = 0
a = 1; b = - 1; c = - 2
Δ = b² - 4ac
Δ = (-1)² - 4.1.(-2)
Δ = 1 + 8
Δ = 9
x = - b +/- √Δ = - ( - 1) +/- √9
2a 2.1
x = 1 + 3 = 4/2 = 2
2
x = 1 - 3 = - 2/2 = - 1
2
R.: x = 2 e x = - 1
**********************************************
h)
y= x² - 2x - 3
0 = x² - 2x - 3
x² - 2x - 3 = 0
a = 1; b = - 2; c = - 3
Δ = b² - 4ac
Δ = (-2)² - 4.1.(-3)
Δ = 4 + 12
Δ = 16
x = - b +/- √Δ = - ( - 2) +/- √16
2a 2.1
x = 2 + 4 = 6/2 = 3
2
x = 2 - 4 = - 2/2 = - 1
2