Respostas

  • Usuário do Brainly
2013-05-28T16:23:12-03:00

a) \text{x}^2-9=27

 

\text{x}^2=36

 

\text{x}=\pm\sqrt{36}

 

\text{x}=\pm6

 

 

b) (\text{x}+7)\cdot(-\text{x}+11)=0

 

-\text{x}^2+11\text{x}-7\text{x}+77=0

 

-\text{x}^2+4\text{x}+77=0

 

\text{x}=\dfrac{-4\pm\sqrt{4^2-4\cdot(-1)\cdot77}}{2\cdot(-1)}=\dfrac{-4\pm18}{-2}

 

\text{x}'=\dfrac{-4+18}{-2}=-7

 

\text{x}"=\dfrac{-4-18}{-2}=11

 

 

 

c) 2\text{x}^2+1=0

 

2\text{x}^2=-1

 

\text{x}^2=\dfrac{-1}{2}

 

Desta maneira, não há soluçõe reais.

 

 

d) 3\text{x}^2-12=0

 

3\text{x}^2=12

 

\text{x}^2=4

 

\text{x}=\pm\sqrt{4}

 

\text{x}=\pm2

 

 

 

e) 5\text{x}^2+125=0

 

5\text{x}^2=-125

 

\text{x}^2=-25

 

Desse modo, não existem soluções reais.