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2x^2+12x+3=0

 Oct 24, 2014

Best Answer 

 #1
avatar+128578 
+5

2x^2+12x+3=0

This polynomial can't be factored.....it does have "real" solutions, though...usng the on-site solver, we have

 

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{30}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{30}}}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{5.738\: \!612\: \!787\: \!525\: \!830\: \!6}}\\
{\mathtt{x}} = -{\mathtt{0.261\: \!387\: \!212\: \!474\: \!169\: \!4}}\\
\end{array} \right\}$$

 

 Oct 24, 2014
 #1
avatar+128578 
+5
Best Answer

2x^2+12x+3=0

This polynomial can't be factored.....it does have "real" solutions, though...usng the on-site solver, we have

 

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{30}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{30}}}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{5.738\: \!612\: \!787\: \!525\: \!830\: \!6}}\\
{\mathtt{x}} = -{\mathtt{0.261\: \!387\: \!212\: \!474\: \!169\: \!4}}\\
\end{array} \right\}$$

 

CPhill Oct 24, 2014

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