As raizes zeros da funções a)f(×)=2ײ-3×+1
b)f(×)=-ײ+2×+15 c)f(×)=-ײ+6×-9 d)f(×)=-x²-5×+9 e)f(×)=-ײ-×-6
Soluções para a tarefa
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a) f(x) = 2x² -3 +1
Δ = (-3)² -4.2.1
Δ = 9 -8
Δ = 1
x = - (-3) ± √1
2 . 2
x1 = 3 + 1 = 4 = 1
4 4
x2 = 3 - 1 = 2 = -2
4 4
b) f(×) = -ײ+2×+15
Δ = 2² -4 . -1 . 15
Δ = 4 + 60
Δ = 64
x = -2 ± √64
2 . -1
x1 = -2 + 8 = 6 = -3
-2 -2
x2 = -2 -8 = -10 = 5
-2 -2
c) f(×) = -ײ+6×-9
Δ = 6² -4 . -1 . -9
Δ = 36 - 36
Δ = 0
x = -6 + √0 = -6 = 3
2 . -1 -2
d) f(×) = -x²-5×+9
Δ = (-5)² -4 . -1 . 9
Δ = 25 +36
Δ = 61
∉ lR
e) f(×) = -ײ-×-6
Δ = (-1)² -4 . -1 . -6
Δ = 1 -24
Δ = -23
∉ lR
Δ = (-3)² -4.2.1
Δ = 9 -8
Δ = 1
x = - (-3) ± √1
2 . 2
x1 = 3 + 1 = 4 = 1
4 4
x2 = 3 - 1 = 2 = -2
4 4
b) f(×) = -ײ+2×+15
Δ = 2² -4 . -1 . 15
Δ = 4 + 60
Δ = 64
x = -2 ± √64
2 . -1
x1 = -2 + 8 = 6 = -3
-2 -2
x2 = -2 -8 = -10 = 5
-2 -2
c) f(×) = -ײ+6×-9
Δ = 6² -4 . -1 . -9
Δ = 36 - 36
Δ = 0
x = -6 + √0 = -6 = 3
2 . -1 -2
d) f(×) = -x²-5×+9
Δ = (-5)² -4 . -1 . 9
Δ = 25 +36
Δ = 61
∉ lR
e) f(×) = -ײ-×-6
Δ = (-1)² -4 . -1 . -6
Δ = 1 -24
Δ = -23
∉ lR
eliveltonv38:
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