Question

ME AJUDEM PF!!
CÁLCULOS
1. f(x) = x² + 6x +8

2. f(x) = - x² - 2x + 15

3. f(x) = 2x² - 8x + 8

4. f(x) = x² - 8x + 12

5. f(x) = x² + 2x - 8​

Answer 1

Resposta:

Explicação passo-a-passo:

1) x^2 + 6x + 8 = 0

Δ= (6)^2 -4.1.8

Δ= 36-32=4

x' = -6 ±$\sqrt{4}$ / 2

x' = -4+2/2

x' = -2/2 = -1

x'' = -4-2/2

x'' = -6/2 = -3

x'= -1 e x'' = -3

2) -x^2 -2x + 15=0

Δ= (-2)^2 -4.(-1).15

Δ= 4 + 60 = 64

x' = -(-2) ± $\sqrt{64}$ / 2.(-1)

x' = 4+8/-2

x'= -12/2 = -6

x'' = 4-8/-2

x'' = -4/-2 = 2

x'= -6 e x''= 2

3) 2x^2 -8x + 8 (vou simplificar todo mundo por 2 para a conta ficar menor)

x^2 -4x +4 = 0

Δ= (-4)^2 -4.1.4

Δ= 16-16 = 0

x' = -(-4) ± $\sqrt{0}$ /2

x' = 4/2 = 2

x'' 4/2= 2

x' = 2 e x'' = 2

4) x^2 -8x +12 = 0

Δ= (-8)^2 -4.1.12

Δ= 64 -48 = 16

x'= -(-8) ± $\sqrt{16}$ /2

x' = 8+4/2

x= 12/2 = 6

x'' = 8-4/2

x'' = 4/2 = 2

x'= 6 e x''=2

5) x^2 +2x -8=0

Δ= (2)^2 -4.1.(-8)

Δ= 4+32=36

x'= -2± $\sqrt{36}$ /2

x' = -2+6/2

x' = 4/2 = 2

x'' = -2-6/2

x'' = -8/2 = -4

x' = 2 e x'' = -4