Question

dertermine as seguintes equações de 2 grau
a)x²-6x-16=0
b)6x²-x-5=0
ajuda pfvr compensação de falta

Answer 1

Resposta:

a) x² - 6x - 16 = 0

a = 1, b = - 6, c ,= - 16

∆ = b² - 4.a.c

∆ = (-6)² - 4.1.(-16)

∆ = 36 - 4.(-16)

∆ = 36 + 64

∆ = 100

x = - b ± √∆ / 2.a

x = -(-6) ± √100 / 2.1

x = 6 ± 10 / 2

x' = 6 + 10 = 16 ÷ 2 = 8

x'' = 6 - 10 = (-4) ÷ 2 = - 2

x' = 8

x'' = - 2

b) 6x² - x - 5 = 0

a = 6, b = - 1, c = - 5

∆ = b² - 4.a.c

∆ = (-1)² - 4.6.(-5)

∆ = 1 - 4.(-30)

∆ = 1 + 120

∆ = 121

x = - b ± √∆ / 2.a

x = -(-1) ± √∆ / 2.6

x = 1 ± 11 / 12

x' = 1 + 11 = 12 ÷ 12 = 1

x" = 1 - 11 = -10/12 = -5/6

x' = 1

x" = -5/6

Answer 2

A

$\large\boxed{\begin{array}{lr}x {}^{2} - 6x - 16 = 0 \\ \left\{\begin{array}{ll}a = 1 \\ b = - 6 \\ c = - 16\end{array}\right. \\ \\ ∆ = b {}^{2} - 4.a.c \\ ∆ = ( - 6) {}^{2} - 4.1.( - 16) \\∆ = 36 + 64 \\ ∆ = 100 \\ \\ x = \frac{ - b \pm \sqrt{∆} }{2.a} \\ \\ x = \frac{ - ( - 6) \pm \sqrt{100} }{2.1} \\ \\ x = \frac{6 \pm10}{2} \\ \\ x_{1} = \frac{6 + 10}{2} = \frac{16}{2} =8 \\ \\ x_{2} = \frac{6 - 10}{2} = \frac{ - 4}{2} = - 2\end{array}}$

B

$\large\boxed{\begin{array}{lr} {6x}^{2} - x - 5 = 0 \\ \left\{\begin{array}{ll}a = 6 \\ b = - 1 \\ c = - 5 \end{array}\right. \\ \\ ∆ = b {}^{2} - 4.a.c \\ ∆ = ( - 1) {}^{2} - 4.6.( - 5) \\∆ = 1 + 120 \\ ∆ = 121 \\ \\ x = \frac{ - b \pm \sqrt{∆} }{2.a} \\ \\ x = \frac{ - ( - 1) \pm \sqrt{121} }{2.6} \\ \\ x = \frac{1 \pm11}{12} \\ \\ x_{1} = \frac{1 + 11}{12} = \frac{12}{12} = 1 \\ \\ x_{2} = \frac{1 - 11}{12} = - \frac{10}{12}=- \frac{5}{6} \end{array}}$