Respostas

2013-06-19T15:06:39-03:00

x² -2x -10=0

 

-b+\frac{-b+- \sqrt{b^{2} -4.a.c}}{2.a}

 

respostas:

\frac{-2+2\sqrt{10}}{2}

e

\frac{-2-2\sqrt{10}}{2}

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2x(4x-7)=8

8x²-14x -8=0

 

-b+\frac{-b+- \sqrt{b^{2} -4.a.c}}{2.a}

 

resposta

 

\frac{7+-\sqrt{113}}{8}

  • Usuário do Brainly
2013-06-19T15:41:17-03:00

a) \text{x}^2-2\text{x}-10=0

 

\text{x}=\dfrac{-\text{b}\pm\sqrt{\Delta}}{2\cdot\text{a}}

 

Onde, \Delta=\text{b}^2-4\cdot\text{a}\cdot\text{c}

 

Segundo o enunciado, \text{a}=1, \text{b}=-2 e \text{c}=-10.

 

Desta maneira, temos \Delta=(-2)^2-4\cdot1\cdot(-10)=4+40=44

 

Logo, \text{x}=\dfrac{-(-2)\pm\sqrt{44}}{2\cdot1}=\dfrac{2\pm2\sqrt{11}}{2}

 

As raízes são:

 

\text{x}'=\dfrac{2+2\sqrt{10}}{2}=1+\sqrt{11}

 

\text{x}"=\dfrac{2-2\sqrt{10}}{2}=1-\sqrt{11}

 

 

b) 2\text{x}\cdot(4\text{x}-7)=8

 

8\text{x}^2-14\text{x}-8=0 

 

\text{x}=\dfrac{-(-14)\pm\sqrt{(-14)^2-4\cdot8\cdot(-8)}}{2\cdot8}=\dfrac{14\pm2\sqrt{113}}{2\cdot8}

 

\text{x}'=\dfrac{14+2\sqrt{113}}{8}=\dfrac{7+\sqrt{113}}{4}

 

\text{x}"=\dfrac{14-2\sqrt{113}}{8}=\dfrac{7-\sqrt{113}}{4}