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

2 - Faça a fatoração e encontre as raízes das equações: a) -3x² + 9x = 0 b) 2x² + 3x = 0 c) (x+5)(x-1) = 2x - 5 d) x 2 + 3x + 2 = 0 e) x 2 + 6x + 8 = 0 f) x 2 - 2x - 15 = 0 g) (x + 1)(x - 3) = x + 7

Respostas

respondido por: Anônimo
1

Explicação passo-a-passo:

a) \sf -3x^2+9x=0

\sf -3x\cdot(x-3)=0

\sf -3x=0~\rightarrow~x'=0

\sf x-3=0~\rightarrow~x"=3

b) \sf 2x^2+3x=0

\sf x\cdot(2x+3)=0

\sf x'=0

\sf 2x+3=0~\rightarrow~x"=\dfrac{-3}{2}

c) \sf (x+5)\cdot(x-1)=2x-5

\sf x^2-x+5x-5-2x+5=0

\sf x^2+2x=0

\sf x\cdot(x+2)=0

\sf x'=0

\sf x+2=0~\rightarrow~x"=-2

d) \sf x^2+3x+2=0

\sf x^2+x+2x+2=0

\sf x\cdot(x+1)+2\cdot(x+1)=0

\sf (x+1)\cdot(x+2)=0

\sf x+1=0~\rightarrow~x'=-1

\sf x+2=0~\rightarrow~x"=-2

e) \sf x^2+6x+8=0

\sf x^2+2x+4x+8=0

\sf x\cdot(x+2)+4\cdot(x+2)=0

\sf (x+2)\cdot(x+4)=0

\sf x+2=0~\rightarrow~x'=-2

\sf x+4=0~\rightarrow~x"=-4

f) \sf x^2-2x-15=0

\sf x^2+3x-5x-15=0

\sf x\cdot(x+3)-5\cdot(x+3)=0

\sf (x+3)\cdot(x-5)=0

\sf x+3=0~\rightarrow~x'=-3

\sf x-5=0~\rightarrow~x"=5

g) \sf (x+1)\cdot(x-3)=x+7

\sf x^2-3x+x-3-x-7=0

\sf x^2-3x-10=0

\sf x^2+2x-5x-10=0

\sf x\cdot(x+2)-5\cdot(x+2)=0

\sf (x+2)\cdot(x-5)=0

\sf x+2=0~\rightarrow~x'=-2

\sf x-5=0~\rightarrow~x"=5

respondido por: Makaveli1996
1

Oie, Td Bom?!

a)

 - 3x {}^{2}  + 9x = 0 \\  - 3x \: . \: (x - 3) = 0 \\ x \: . \: (x - 3) = 0

x = 0 \\ x - 3  = 0⟶x = 3

 S= \left \{  x _{1} = 0 \:  \: x_{2} = 3 \right \}

b)

2x {}^{2}  + 3x = 0 \\ x \: . \: (2x + 3) = 0

x = 0 \\ 2x + 3 = 0⟶x =  -  \frac{3}{2}

S = \left \{ x_{1} = 0 \:  \: x_{2} =  -  \frac{3}{2}   \right \}

c)

(x + 5) \: . \: (x - 1) = 2x - 5 \\ x {}^{2}  - x + 5x - 5 = 2x - 5 \\   x{}^{2}  - x + 5x = 2x \\ x {}^{2}  + 4x = 2x \\ x {}^{2}  + 4x - 2x = 0 \\ x {}^{2}  + 2x = 0 \\ x \: . \: (x + 2) = 0

x = 0 \\ x + 2 = 0⟶x =  - 2

 S= \left \{x_{1} = 0 \:  \:  x_{2} =  - 2   \right \}

d)

x {}^{2}  + 3x + 2 = 0 \\ x {}^{2}  + 2x + x + 2 = 0 \\ x \: . \: (x + 2) + x + 2 = 0 \\ (x + 2) \: . \: (x +1 ) = 0

x + 2 = 0 ⟶x =  - 2\\ x + 1 = 0⟶x =  - 1

S = \left \{x_{1} =  - 2 \:  \:  x_{2} =  - 1   \right \}

e)

x {}^{2}  + 6x + 8 = 0 \\ x {}^{2}  + 4x + 2x + 8 = 0 \\ x \: . \: (x + 4) + 2(x + 4) = 0 \\ (x + 4) \: . \: (x + 2) = 0

x + 4 = 0 ⟶x =  - 4\\ x + 2 = 0⟶x =  - 2

 S= \left \{x_{1} =  - 4 \:  \:  x_{2} =  - 2  \right \}

f)

 x {}^{2}  - 2x - 15 = 0\\ x {}^{2}  + 3x - 5x - 15  = 0 \\ x \: . \: (x + 3) - 5(x + 3) = 0 \\ (x + 3) \: . \: (x - 5) = 0

x + 3 = 0⟶x =  - 3 \\ x - 5 = 0⟶x = 5

 S= \left \{ x_{1} =  - 3 \:  \: x_{2} = 5  \right \}

g)

(x + 1) \: . \: (x - 3) = x + 7 \\ x {}^{2}  - 3x + x - 3 = x + 7 \\ x {}^{2}  - 3x - 3 = 7 \\ x {}^{2}  - 3x - 3 - 7 = 0 \\ x {}^{2}  + 2x - 5x - 10 = 0 \\ x \: . \: (x + 2) - 5(x + 2) = 0 \\ (x + 2) \: . \: (x - 5) = 0

x + 2 = 0⟶x =  - 2 \\ x - 5 = 0⟶x = 5

 S= \left \{x_{1} =  - 2 \:  \:  x_{2} = 5  \right \}

Att. Makaveli1996

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