I have been reading the compleate works of Lewis Carrol. A lot of people don't realise he was also a well respected mathematician. I came accross the equation x^2+7x+53=11/3. The first time I entered it into this calculator, it came up with a solution with imaginary numbers. But now, the second time, it comes up with a compleatly diffrent answer. I do not know if this is the calculator's falt, or if the equation is so nonsensical that it doesn't have one answer. Could someone please tell me an approxomate answer? It would be greatly appriciated. I know there is no real number answer fo this, but if i could have the closest possible answer in imaginary numbers, I would greatly appriciate it. Thank you.
+130511 Well let's see...mutiply both sides by 3
3x^2 + 21x +159 = 11 Subtract 11 from both sides
3x^2 + 21x + 148 = 0
I can tell right away that this has no "real" solutions....Let's see what he "magic" calculator sez.......
$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{148}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{1\,335}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}\right)}{{\mathtt{6}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{1\,335}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{21}}\right)}{{\mathtt{6}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{7}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6.089\: \!608\: \!635\: \!484\: \!241\: \!2}}{i}\right)\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{7}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6.089\: \!608\: \!635\: \!484\: \!241\: \!2}}{i}\\
\end{array} \right\}$$
Yep...we get two "complex" solutions...as expected....
+3502 the numbers were suppose to stand for a riddle plus you cannot find an answer if this were a real problem because the equation has already been solved within itself x^2+7x+53=11/3.
+130511 Well let's see...mutiply both sides by 3
3x^2 + 21x +159 = 11 Subtract 11 from both sides
3x^2 + 21x + 148 = 0
I can tell right away that this has no "real" solutions....Let's see what he "magic" calculator sez.......
$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{148}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{1\,335}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}\right)}{{\mathtt{6}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{1\,335}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{21}}\right)}{{\mathtt{6}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{7}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6.089\: \!608\: \!635\: \!484\: \!241\: \!2}}{i}\right)\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{7}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6.089\: \!608\: \!635\: \!484\: \!241\: \!2}}{i}\\
\end{array} \right\}$$
Yep...we get two "complex" solutions...as expected....
