Resposta :
Opa, beleza colega!?
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Resposta:
A) x - x - 12 = 0
a) 1
a) 1b) 1
a) 1b) 1c) -12
a) 1b) 1c) -12Δ = b² - 4ac
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0Δ = 1 - 0
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0Δ = 1 - 0Δ = 1
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0Δ = 1 - 0Δ = 1x= -b ±√∆
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0Δ = 1 - 0Δ = 1x= -b ±√∆2.a
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0Δ = 1 - 0Δ = 1x= -b ±√∆2.ax= -1 ± 1
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0Δ = 1 - 0Δ = 1x= -b ±√∆2.ax= -1 ± 1x= 1 + 1 = 2/2 = 1
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0Δ = 1 - 0Δ = 1x= -b ±√∆2.ax= -1 ± 1x= 1 + 1 = 2/2 = 1x= 1 - 1 = 0
a) 1b) 1c) -12Δ = b² - 4acΔ = (1)² - 4.(-12).0Δ = 1 - 0Δ = 1x= -b ±√∆2.ax= -1 ± 1x= 1 + 1 = 2/2 = 1x= 1 - 1 = 0{1,0}
B) x² - 4x - 5 = 0
a) 1
a) 1b) -4
a) 1b) -4c) -5
a) 1b) -4c) -5Δ = b² - 4ac
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)Δ = 16 + 20
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)Δ = 16 + 20Δ = 36
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)Δ = 16 + 20Δ = 36x= -b ±√∆
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)Δ = 16 + 20Δ = 36x= -b ±√∆2.a
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)Δ = 16 + 20Δ = 36x= -b ±√∆2.ax= -(-4) ± 6
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)Δ = 16 + 20Δ = 36x= -b ±√∆2.ax= -(-4) ± 6x= 4 + 6 = 10/2 = 5
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)Δ = 16 + 20Δ = 36x= -b ±√∆2.ax= -(-4) ± 6x= 4 + 6 = 10/2 = 5x= 4 - 6 = -2/2 = -1
a) 1b) -4c) -5Δ = b² - 4acΔ = (-4)² - 4.1.(-5)Δ = 16 + 20Δ = 36x= -b ±√∆2.ax= -(-4) ± 6x= 4 + 6 = 10/2 = 5x= 4 - 6 = -2/2 = -1{5, -1}
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•ASHELEY•
Resposta:
a) x'= 4 x''= -3
b) x'= 5 x''= -1
Explicação passo-a-passo:
a)
[tex]x = \frac{ - ( { - 1)}^{2} \frac{ + }{ - } \sqrt{ { (- 1)}^{2} - 4 \times (1) \times ( - 12)} }{2 \times (1)} \\ x = \frac{1 \frac{ + }{ - } \sqrt{1 + 48} }{2} \\ x = \frac{1 \frac{ + }{ - } \sqrt{49} }{2} \\ x = \frac{1 \frac{ + }{ - } 7}{2} \\ \\ x1 = \frac{8}{2} = 4 \: \: \: \: \: \: \: \: \: x2 = \frac{ - 6}{2} = - 3[/tex]
b)
[tex]x = \frac{ - ( - 4) \frac{ + }{ - } \sqrt{ {( - 4)}^{2} - 4 \times (1) \times ( - 5) } }{2 \times (1)} \\ x = \frac{4 \frac{ + }{ - } \sqrt{16 + 20} }{2} \\ x = \frac{4 \frac{ + }{ - } \sqrt{36} }{2} \\ x = \frac{4 \frac{ + }{ - } 6}{2} \\ \\ x1 = \frac{10}{2} = 5 \: \: \: \: \: \: \: \: \: x2 = \frac{ - 2}{2} = - 1[/tex]