Respostas

  • Usuário do Brainly
2013-02-05T00:33:24-02:00

a)

 

-\text{x}^2+\text{x}+12=0

 

\text{x}=\dfrac{-1\pm\sqrt{1^2-4\cdot(-1)\cdot12}}{2\cdot-1}=\dfrac{-1\pm7}{-2}

 

\text{x}'=\dfrac{-1+7}{-2}=-3

 

\text{x}"=\dfrac{-1-7}{-2}=4

 

\text{S}=\{-3, 4\}

 

b)

 

4\text{x}^2-\text{x}+1=\text{x}+3\text{x}^2

 

\text{x}^2-2\text{x}+1=0

 

\text{x}=\dfrac{-(-2)\pm\sqrt{(-2)^2-4\cdot1\cdot1}}{2\cdot1}=\dfrac{2\pm0}{2}

 

\text{x}'=\text{x}"=\dfrac{2\pm0}{2}=1

 

\text{S}=\{1\}

 

c)

 

\text{x}^2-2\text{x}+4=0

 

\text{x}=\dfrac{-(-2)\pm\sqrt{(-2)^2-4\cdot1\cdot4}}{2\cdot1}=\dfrac{2\pm\sqrt{-12}}{2}

 

Como \Delta<0 não há soluções reais.

 

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