I got three questions that I am stuck on and not sure how it works.
the first question is what is the approximate values of x that are solutions to f(x)=0, where f(x)=-8x^2+4x+3 the options for this answer are {-2.67,1.33} , {-8,4} , {-0.50,-0.38} , {-0.41,0.91}
second question is same, only now it is f(x)=-4x^2+8x+5 . the options for this answer are {2.50,-0.50} , {-4,8} , {-2.00,-1.25} , {-0.80,1.60}
and the same for the third question, but now it is f(x)=-9x^2+7x+6 . the options for this answer are {-0.52,1.29} , {-9,7} , {-0.78,-0.67} , {-1.50,1.17}
Thank you!
+130516
-8x^2 + 4x + 3 = 0 this won't factor....using the onsite solver and the Quadratic Formula, we have
$${\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{7}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.411\: \!437\: \!827\: \!766\: \!147\: \!6}}\\
{\mathtt{x}} = {\mathtt{0.911\: \!437\: \!827\: \!766\: \!147\: \!6}}\\
\end{array} \right\}$$
-4x^2 + 8x + 5 = 0 factor
(-2x -1)(2x -5) = 0 and setting each factor to 0, we have that x = -0.5 and x = 2.5
-9x^2 + 7x + 6 = 0 same as the first......no factorization is possible
$${\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{265}}}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{18}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{265}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{18}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.515\: \!490\: \!033\: \!116\: \!650\: \!4}}\\
{\mathtt{x}} = {\mathtt{1.293\: \!267\: \!810\: \!894\: \!428\: \!1}}\\
\end{array} \right\}$$
So.....x = -0.52 and x = 1.29
+130516
-8x^2 + 4x + 3 = 0 this won't factor....using the onsite solver and the Quadratic Formula, we have
$${\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{7}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.411\: \!437\: \!827\: \!766\: \!147\: \!6}}\\
{\mathtt{x}} = {\mathtt{0.911\: \!437\: \!827\: \!766\: \!147\: \!6}}\\
\end{array} \right\}$$
-4x^2 + 8x + 5 = 0 factor
(-2x -1)(2x -5) = 0 and setting each factor to 0, we have that x = -0.5 and x = 2.5
-9x^2 + 7x + 6 = 0 same as the first......no factorization is possible
$${\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{265}}}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{18}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{265}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{18}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.515\: \!490\: \!033\: \!116\: \!650\: \!4}}\\
{\mathtt{x}} = {\mathtt{1.293\: \!267\: \!810\: \!894\: \!428\: \!1}}\\
\end{array} \right\}$$
So.....x = -0.52 and x = 1.29
