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What are the approximate values of x that are solutions to f(x) = 0, where f(x) = -9x2 + 5x + 3.

 Mar 1, 2015

Best Answer 

 #1
avatar+130516 
+5

 

 

 

Here are the values....

$${\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{133}}}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}{{\mathtt{18}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{133}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}{{\mathtt{18}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.362\: \!920\: \!144\: \!148\: \!377\: \!5}}\\
{\mathtt{x}} = {\mathtt{0.918\: \!475\: \!699\: \!703\: \!933\: \!1}}\\
\end{array} \right\}$$

 

 

 Mar 1, 2015
 #1
avatar+130516 
+5
Best Answer

 

 

 

Here are the values....

$${\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{133}}}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}{{\mathtt{18}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{133}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}{{\mathtt{18}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.362\: \!920\: \!144\: \!148\: \!377\: \!5}}\\
{\mathtt{x}} = {\mathtt{0.918\: \!475\: \!699\: \!703\: \!933\: \!1}}\\
\end{array} \right\}$$

 

 

CPhill Mar 1, 2015

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