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x^4-15x^2+64=0

 Jun 22, 2015

Best Answer 

 #1
avatar+33603 
+5

There are no real number solutions to this, just complex ones:

 

$${{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{64}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\sqrt{{\mathtt{31}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\sqrt{{\mathtt{31}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{15}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{31}}}}{\mathtt{\,\times\,}}{i}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{15}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{31}}}}{\mathtt{\,\times\,}}{i}}}}{{\sqrt{{\mathtt{2}}}}}}\\
\end{array} \right\}$$

.

 Jun 23, 2015
 #1
avatar+33603 
+5
Best Answer

There are no real number solutions to this, just complex ones:

 

$${{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{64}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\sqrt{{\mathtt{31}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\sqrt{{\mathtt{31}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{15}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{31}}}}{\mathtt{\,\times\,}}{i}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{15}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{31}}}}{\mathtt{\,\times\,}}{i}}}}{{\sqrt{{\mathtt{2}}}}}}\\
\end{array} \right\}$$

.

Alan Jun 23, 2015

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