The sum of two numbers is 6, and the sum of their squares is 28. Find these numbers exactly

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Answer to a math question The sum of two numbers is 6, and the sum of their squares is 28. Find these numbers exactly

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$\left\{\begin{array} { l } x+y=6 \\ {x}^{2}+{y}^{2}=28\end{array} \right.$

$\left\{\begin{array} { l } x=6-y \\ {x}^{2}+{y}^{2}=28\end{array} \right.$

${\left( 6-y \right)}^{2}+{y}^{2}=28$

$\begin{array} { l }y=3+\sqrt{ 5 },\\y=3-\sqrt{ 5 }\end{array}$

$\begin{array} { l }x=6-\left( 3+\sqrt{ 5 } \right),\\y=3-\sqrt{ 5 }\end{array}$

$\begin{array} { l }x=6-\left( 3+\sqrt{ 5 } \right),\\x=6-\left( 3-\sqrt{ 5 } \right)\end{array}$

$\begin{array} { l }x=3-\sqrt{ 5 },\\x=6-\left( 3-\sqrt{ 5 } \right)\end{array}$

$\begin{array} { l }x=3-\sqrt{ 5 },\\x=3+\sqrt{ 5 }\end{array}$

$\begin{array} { l }\left( x_1, y_1\right)=\left( 3-\sqrt{ 5 }, 3+\sqrt{ 5 }\right),\\\left( x_2, y_2\right)=\left( 3+\sqrt{ 5 }, 3-\sqrt{ 5 }\right)\end{array}$

$\begin{array} { l }\left\{\begin{array} { l } 3-\sqrt{ 5 }+3+\sqrt{ 5 }=6 \\ {\left( 3-\sqrt{ 5 } \right)}^{2}+{\left( 3+\sqrt{ 5 } \right)}^{2}=28\end{array} \right.,\\\left\{\begin{array} { l } 3+\sqrt{ 5 }+3-\sqrt{ 5 }=6 \\ {\left( 3+\sqrt{ 5 } \right)}^{2}+{\left( 3-\sqrt{ 5 } \right)}^{2}=28\end{array} \right.\end{array}$

$\begin{array} { l }\left\{\begin{array} { l } 6=6 \\ 28=28\end{array} \right.,\\\left\{\begin{array} { l } 3+\sqrt{ 5 }+3-\sqrt{ 5 }=6 \\ {\left( 3+\sqrt{ 5 } \right)}^{2}+{\left( 3-\sqrt{ 5 } \right)}^{2}=28\end{array} \right.\end{array}$

$\begin{array} { l }\left\{\begin{array} { l } 6=6 \\ 28=28\end{array} \right.,\\\left\{\begin{array} { l } 6=6 \\ 28=28\end{array} \right.\end{array}$

$\begin{array} { l }\left( x_1, y_1\right)=\left( 3-\sqrt{ 5 }, 3+\sqrt{ 5 }\right),\\\left( x_2, y_2\right)=\left( 3+\sqrt{ 5 }, 3-\sqrt{ 5 }\right)\end{array}$

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