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Soluções para a tarefa
Explicação passo-a-passo:
1)
x² + 5x + 4 = 0
a= 1; b = 5; c = 4
D = 5² - 4 . 1 . 4
D = 25 - 16
D = 9
x' = -5 + 3
2 . 1
x' = -5 + 3
2
x' = -2
2
x' = -1
x" = -5 - 3
2 . 1
x" = -5 - 3
2
x'' = -8
2
x'' = -4
2)
x² - 6x + 9 = 0
a= 1; b = -6; c = 9
D = -6² - 4 . 1 . 9
D = 36 - 36
D = 0
x' = -(-6) + 0
2 . 1
x' = 6 + 0
2
x' = 6
2
x' = 3
x" = -(-6) - 0
2 . 1
x" = 6 - 0
2
x'' = 6
2
x'' = 3
3)
f(-2) = 1.(-2)² - 1. (-2) + 0 = 1.4 + 2 + 0 = 6
f(0) = 1.(0)² - 1.0 + 0 = 1.0 + 0 + 0 = 0
f(4) = 1.(4)² - 1.4 + 0 = 1.16 - 4 + 0 = 12
f(-3) = 1.(-3)² - 1. (-3) + 0 = 1.9 + 3 + 0 = 12
6 + 12 + 0 - 12 = 6
4)
a)
x² - 2x + 1 = 0
a= 1; b = -2; c = 1
D = -2² - 4 . 1 . 1
D = 4 - 4
D = 0
x' = -(-2) + 0
2 . 1
x' = 2 + 0
2
x' = 2
2
x' = 1
x" = -(-2) - 0
2 . 1
x" = 2 - 0
2
x'' = 2
2
x'' = 1
b)
x² - x - 2 = 0
a= 1; b = -1; c = -2
D = -1² - 4 . 1 . (-2)
D = 1 + 8
D = 9
x' = -(-1) + 3
2 . 1
x' = 1 + 3
2
x' = 4
2
x' = 2
x" = -(-1) - 3
2 . 1
x" = 1 - 3
2
x'' = -2
2
x'' = -1
c)
2x² + 3x + 4 = 0
a= 2; b = 3; c = 4
D = 3² - 4 . 2 . 4
D = 9 - 32
D = -23