Question

3-Das equaçoes da questão anterior resolva as que tem solucao usando a formula resolutiva de Baskhara.

a)x^2 -2x+1=0

b)3x^2 -2x-1=0

c)x^2 +3x-10=0

d)x^2 -6x+9=0

Obs:As q tem solucao da questao anterior​

Answer 1

a) x^2 - 2x + 1 = 0

a = 1

b = -2

c = 1

Δ = b^2 - 4 . a . c

Δ = (-2)^2 - 4 . 1 . 1

Δ = 4 - 4

Δ = 0

x = -b ± √Δ /2 . a

x = -(-2) ± √0 /2 . 1

x = 2 ± 0 /2

x' = 2 + 0 /2

x' = 2/2

x' = 1

x" = 2 - 0 /2

x" = 2/2

x" = 1

S = {1 , 1}

b) 3x^2 - 2x - 1 = 0

a = 3

b = -2

c = -1

Δ = b^2 - 4 . a . c

Δ = (-2)^2 - 4 . 3 . (-1)

Δ = 4 - (-12)

Δ = 4 + 12

Δ = 16

x = -b ± √Δ /2 . a

x = -(-2) ± √16 /2 . 3

x = 2 ± 4 /6

x' = 2 + 4 /6

x' = 6/6

x' = 1

x" = 2 - 4 /6

x" = -2/6

x" = -1/3

S = {1 , -1/3}

c) x^2 + 3x - 10 = 0

a = 1

b = 3

c = -10

Δ = b^2 - 4 . a . c

Δ = 3^2 - 4 . 1 . (-10)

Δ = 9 - (-40)

Δ = 9 + 40

Δ = 49

x = -b ± √Δ /2 . a

x = -3 ± √49 /2 . 1

x = -3 ± 7 /2

x' = -3 + 7 /2

x' = 4/2

x' = 2

x" = -3 - 7 /2

x" = -10/2

x" = -5

S = {2 , -5}

d) x^2 - 6x + 9 = 0

a = 1

b = -6

c = 9

Δ = b^2 - 4 . a . c

Δ = (-6)^2 - 4 . 1 . 9

Δ = 36 - 4 . 9

Δ = 36 - 36

Δ = 0

x = -b ± √Δ /2 . a

x = -(-6) ± √0 /2 . 1

x = 6 ± 0 /2

x' = 6 + 0 /2

x' = 6/2

x' = 3

x" = 6 - 0 /2

x" = 6/2

x" = 3

S = {3 , 3}