Respostas

2013-07-31T21:44:32-03:00
1) Pegando a equação:

2x^{2}+14x+2=0

Dividindo-a toda por 2 e resolvendo-a, temos:

x^{2}+7x+1=0

\Delta=b^{2}-4\cdot a\cdot c
\Delta=7^{2}-4\cdot1\cdot1
\Delta=49-4
\Delta=45

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

x=\dfrac{-1\pm\sqrt{45}}{2\cdot1}

x=\dfrac{-1\pm3\sqrt{5}}{2}

S=\{\dfrac{-1-3\sqrt{5}}{2},\;\dfrac{-1+3\sqrt{5}}{2}\}

2) Resolvendo a equação, temos:

2x^{2}+1x-3=0

\Delta=b^{2}-4\cdot a\cdot c
\Delta=1^{2}-4\cdot2\cdot(-3)
\Delta=1+24
\Delta=25

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

x=\dfrac{-1\pm\sqrt{25}}{2\cdot2}

x=\dfrac{-1\pm5}{4}

\Longrightarrow\;x_{1}=\dfrac{-1+5}{4}=\dfrac{4}{4}=1

\Longrightarrow\;x_{2}=\dfrac{-1-5}{4}=\dfrac{-6}{4}=\dfrac{-3}{2}

S=\{\dfrac{-3}{2},\;1\}