• Matéria: Matemática
  • Autor: Anônimo
  • Perguntado 6 anos atrás

10. Determine as raízes reais das equações incompletas: a) x2 − 6x = 0 b) −2x2 − 42x = 0 c) 5x2 + 20x = 0 d) x2 − 7x = 0 e) 10x2 − 90 = 0 f) 25x2 − 1 = 0 g) x2 − 64 = 0 h) x2 + 16 = 0 i) −7x2 + 28 = 0 j) (x − 7)(x − 3) + 10x = 30 k) 2x(x + 1) = x(x + 5) + 3(12 − x)

Respostas

respondido por: Makaveli1996
1

Oie, Td Bom?!

a)

x {}^{2}  - 6x = 0

x \: . \: (x - 6) = 0

x = 0

x - 6 = 0⇒x = 6

S = \left \{  0 \: , \: 6\right \}

b)

 - 2x {}^{2}  - 42x = 0

 - 2x \: . \: (x + 21) = 0

x \: . \: (x + 21) = 0

x = 0

x + 21 = 0⇒ x =  - 21

S = \left \{  - 21 \: , \: 0 \right \}

c)

5x {}^{2}  + 20x = 0

5x \: . \: (x + 4) = 0

x \: . \: (x + 4) = 0

x = 0

x + 4 = 0⇒x =  - 4

S = \left \{  - 4 \: , \: 0 \right \}

d)

x {}^{2}  - 7x = 0

x \: . \: (x - 7) = 0

x = 0

x - 7 = 0⇒x = 7

S = \left \{  0 \: , \: 7\right \}

e)

10x {}^{2}  - 90 = 0

x {}^{2}  - 9 = 0

x {}^{2}  = 9

x = ± \sqrt{9}

x = ±3

S = \left \{   - 3 \: , \: 3\right \}

f)

25x {}^{2}  - 1 = 0

25x {}^{2}  = 1

x {}^{2}  =  \frac{1}{25}

x = ± \sqrt{ \frac{1}{25} }

x = ± \frac{1}{5}

S = \left \{  -  \frac{1}{5}  \:  ,\:  \frac{1}{5}  \right \}

g)

x {}^{2}  - 64 = 0

x {}^{2}  = 64

x = ± \sqrt{64}

x = ±8

S = \left \{  - 8 \: , \: 8 \right \}

h)

x {}^{2}  + 16 = 0

x {}^{2}  =  - 16

x = ± \sqrt{ - 16}

x∉\mathbb{R}

  • A raiz quadrada do número negativo não pertence ao intervalo dos Números Reais.

i)

 - 7x {}^{2}  + 28 = 0

x {}^{2}  - 4 = 0

x {}^{2}  = 4

x = ± \sqrt{4}

x = ±2

S = \left \{   - 2 \:  ,\: 2\right \}

j)

(x - 7) \: . \: (x - 3) + 10x = 30

x {}^{2}  - 3x - 7x + 21 + 10x = 30

x {}^{2}  + 0 + 21 = 30

x {}^{2}  + 21 = 30

x {}^{2}  = 30 - 21

x {}^{2}  = 9

x = ± \sqrt{9}

x = ±3

S = \left \{   - 3 \: , \: 3\right \}

k)

2x \: . \: (x + 1) = x \: . \: (x + 5) + 3(12 - x)

2x {}^{2}  + 2x = x {}^{2}  + 5x + 36 - 3x

2x {}^{2}  + 2x = x {}^{2}  + 2x + 36

2x {}^{2}  = x {}^{2}  + 36

2x {}^{2}  - x {}^{2}  = 36

x {}^{2}  = 36

x = ± \sqrt{36}

x = ±6

S = \left \{  - 6 \: , \: 6 \right \}

Att. Makaveli1996

respondido por: Anônimo
1

Explicação passo-a-passo:

a)

x² - 6x = 0

x.(x - 6) = 0

• x' = 0

• x - 6 = 0

x" = 6

b)

-2x² - 42x = 0

2x² + 42x = 0

2x.(x + 21) = 0

• 2x = 0

x = 0/2

x' = 0

• x + 21 = 0

x" = -21

c)

5x² + 20x = 0

5x.(x + 4) = 0

• 5x = 0

x = 0/5

x' = 0

• x + 4 = 0

x" = -4

d)

x² - 7x = 0

x.(x - 7) = 0

• x' = 0

• x - 7 = 0

x" = 7

e)

10x² - 90 = 0

10x² = 90

x² = 90/10

x² = 9

x = ±√9

• x' = 3

• x" = -3

f)

25x² - 1 = 0

25x² = 1

x² = 1/25

x = ±√1/25

x' = 1/5

x" = -1/5

g)

x² - 64 = 0

x² = 64

x = ±√64

• x' = 8

• x" = -8

h)

x² + 16 = 0

x² = -16

Não há raízes reais

i)

-7x² + 28 = 0

7x² = 28

x² = 28/7

x² = 4

x = ±√4

• x' = 2

• x" = -2

j)

(x - 7).(x - 3) + 10x = 30

x² - 10x + 21 + 10x = 30

x² = 9

x = ±√9

• x' = 3

• x" = -3

k)

2x.(x + 1) = x.(x + 5) + 3.(12 - x)

2x² + 2x = x² + 5x + 36 - 3x

x² = 36

x = ±√36

• x' = 6

• x" = -6

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