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# 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

Sigrid
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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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