Question

Resolva as equações numéricos

A) x2 + x = 0
B) t2 + 5t + 6 = 0
C) x2 -x -2 = 0
D) x(x+2) = 4x

gente alguém pd me ajuda pfv preciso tbm do resultado do
${ x}^{1} x^{2}$
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Answer 1

Explicação passo-a-passo:

Explicação passo-a-passo:

A) x² + x = 0

a = 1

b = 1

c = 0

Δ = b² - 4ac

Δ = (1)² - 4.1.0

Δ = 1 - 0

Δ = 1

x = -b ± √Δ

-----------

2a

x = - 1 ± √1

------------

2.1

x = -1 ± 1

---------

2

x' = -1 + 1

-------

2

x' = 0/2

x' = 0

x'' = - 1 - 1

-------

2

x'' = -2/2

x'' = - 1

S = {0, -1}

B) t² + 5t + 6 = 0

a = 1

b = 5

c = 6

Δ = b² - 4ac

Δ = (5)² - 4.1.6

Δ = 25 - 24

Δ = 1

x = -b ± √Δ

-----------

2a

x = - 5 ± √1

------------

2.1

x = -5 ± 1

---------

2

x' = -5 + 1

-------

2

x' = -4/2

x' = - 2

x'' = - 1 - 5

---------

2

x'' = -6/2

x'' = - 3

S = {-2, -3}

C) x² -x -2 = 0

a = 1

b = -1

c = -2

Δ = b² - 4ac

Δ = (-1)² - 4.1.(-2)

Δ = 1 + 8

Δ = 9

x = -b ± √Δ

-----------

2a

x = - (-1) ± √9

------------

2.1

x = +1 ± 3

---------

2

x' = 1 + 3

-------

2

x' = 4/2

x' = 2

x'' = 1 - 3

---------

2

x'' = -2/2

x'' = - 1

S = {2, -1}

D) x(x+2) = 4x

x² + 2x = 4x

x² + 2x - 4x = 0

x² - 2x = 0

a = 1

b = - 2

c = 0

Δ = b² - 4ac

Δ = (-2)² - 4.1.0

Δ = 4 - 0

Δ = 4

x = -b ± √Δ

-----------

2a

x = - (-2) ± √4

------------

2.1

x = +2 ± 2

---------

2

x' = 2 + 2

-------

2

x' = 4/2

x' = 2

x'' = 2 - 2

---------

2

x'' = 0/2

x'' = 0

S = {2, 0}