Resolva as seguintes equações, utilizando a fórmula resolutiva da equação do 2º grau :

a) x² - x - 2 = 0
x1 =
x2=

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b) x² + 9x + 8 = 0
x1=
x2=

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c) x² - x - 20 = 0
x1=
x2=

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d) x² - 8x + 7 = 0
x1=
x2=

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

e) x² - 3x - 4 = 0
x1=
x2=

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com os calcúlos pra hoje por favor , URGEEEEEEEENTE GENTE :/

1

Respostas

  • Usuário do Brainly
2013-05-28T14:17:42-03:00

a) \text{x}^2-\text{x}-2=0

 

\text{x}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-2)}}{2\cdot1}=\dfrac{1\pm3}{2}

 

\text{x}_1=\dfrac{1+3}{2}=2

 

\text{x}_2=\dfrac{1-3}{2}=-1

 

 

b) \text{x}^2+9\text{x}+8=0

 

\text{x}=\dfrac{-9\pm\sqrt{9^2-4\cdot1\cdot8}}{2\cdot1}=\dfrac{-9\pm7}{2}

 

\text{x}_1=\dfrac{-9+7}{2}=-1

 

\text{x}_2=\dfrac{-9-7}{2}=-8

 

 

c) \text{x}^2-\text{x}-20=0

 

\text{x}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-20)}}{2\cdot1}=\dfrac{1\pm9}{2}

 

\text{x}_1=\dfrac{1+9}{2}=5

 

\text{x}_2=\dfrac{1-9}{2}=-4

 

 

d) \text{x}^2-8\text{x}+7=0

 

\text{x}=\dfrac{-(-8)\pm\sqrt{(-8)^2-4\cdot1\cdot7}}{2\cdot1}=\dfrac{8\pm6}{2}

 

\text{x}_1=\dfrac{8+6}{2}=7

 

\text{x}_2=\dfrac{8-6}{2}=1

 

 

e) \text{x}^2-3\text{x}-4=0

 

\text{x}=\dfrac{-(-3)\pm\sqrt{(-3)^2-4\cdot1\cdot(-4)}}{2\cdot1}=\dfrac{3\pm5}{2}

 

\text{x}_1=\dfrac{3+5}{2}=4

 

\text{x}_2=\dfrac{3-5}{2}=-1

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