Respostas

2013-08-01T17:55:52-03:00
Resolvendo a equação, temos:

x^{2}+12x+36=0

\Delta=b^{2}-4\cdot a\cdot c
\Delta=12^{2}-4\cdot1\cdot36
\Delta=144-144
\Delta=0

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

x=\dfrac{-12\pm\sqrt{0}}{2\cdot1}

x=\dfrac{-12}{2}=-6

\Longrightarrow\;x_{1}=x_{2}=-6

S=\{-6\}
9 4 9
2013-08-01T18:25:46-03:00
x²+12x+36=0
a=1
b=12
c=36
Delta=12²-4*1*36
Delta=144-144
Delta=0

x'=-12+V0/2*1
x'=-12+0/2
x'=-12/2
x'=-6

x''=-12-0/2
x''=-12/2
x''=-6

Solução= {-6}
2 5 2