Find all real numbers x that satisfy the equation \sqrt{x^2-2}=\sqrt{3-x}

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Answer to a math question Find all real numbers x that satisfy the equation \sqrt{x^2-2}=\sqrt{3-x}

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90 Answers

${x}^{2}-2=3-x$

${x}^{2}-2-3+x=0$

${x}^{2}-5+x=0$

${x}^{2}+x-5=0$

$\begin{array} { l }a=1,& b=1,& c=-5\end{array}$

$x=\frac{ -1\pm\sqrt{ {1}^{2}-4 \times 1 \times \left( -5 \right) } }{ 2 \times 1 }$

$x=\frac{ -1\pm\sqrt{ {1}^{2}-4 \times \left( -5 \right) } }{ 2 \times 1 }$

$x=\frac{ -1\pm\sqrt{ {1}^{2}-4 \times \left( -5 \right) } }{ 2 }$

$x=\frac{ -1\pm\sqrt{ 1-4 \times \left( -5 \right) } }{ 2 }$

$x=\frac{ -1\pm\sqrt{ 1+20 } }{ 2 }$

$x=\frac{ -1\pm\sqrt{ 21 } }{ 2 }$

$\begin{array} { l }x=\frac{ -1+\sqrt{ 21 } }{ 2 },\\x=\frac{ -1-\sqrt{ 21 } }{ 2 }\end{array}$

$\begin{array} { l }\sqrt{ {\left( \frac{ -1+\sqrt{ 21 } }{ 2 } \right)}^{2}-2 }=\sqrt{ 3-\frac{ -1+\sqrt{ 21 } }{ 2 } },\\x=\frac{ -1-\sqrt{ 21 } }{ 2 }\end{array}$

$\begin{array} { l }\sqrt{ {\left( \frac{ -1+\sqrt{ 21 } }{ 2 } \right)}^{2}-2 }=\sqrt{ 3-\frac{ -1+\sqrt{ 21 } }{ 2 } },\\\sqrt{ {\left( \frac{ -1-\sqrt{ 21 } }{ 2 } \right)}^{2}-2 }=\sqrt{ 3-\frac{ -1-\sqrt{ 21 } }{ 2 } }\end{array}$

$\begin{array} { l }\frac{ \sqrt{ 14-2\sqrt{ 21 } } }{ 2 }=\frac{ \sqrt{ 14-2\sqrt{ 21 } } }{ 2 },\\\sqrt{ {\left( \frac{ -1-\sqrt{ 21 } }{ 2 } \right)}^{2}-2 }=\sqrt{ 3-\frac{ -1-\sqrt{ 21 } }{ 2 } }\end{array}$

$\begin{array} { l }\frac{ \sqrt{ 14-2\sqrt{ 21 } } }{ 2 }=\frac{ \sqrt{ 14-2\sqrt{ 21 } } }{ 2 },\\\frac{ \sqrt{ 14+2\sqrt{ 21 } } }{ 2 }=\frac{ \sqrt{ 14+2\sqrt{ 21 } } }{ 2 }\end{array}$

$\begin{array} { l }x=\frac{ -1+\sqrt{ 21 } }{ 2 },\\\frac{ \sqrt{ 14+2\sqrt{ 21 } } }{ 2 }=\frac{ \sqrt{ 14+2\sqrt{ 21 } } }{ 2 }\end{array}$

$\begin{array} { l }x=\frac{ -1+\sqrt{ 21 } }{ 2 },\\x=\frac{ -1-\sqrt{ 21 } }{ 2 }\end{array}$

$\begin{align*}&\begin{array} { l }x_1=\frac{ -1-\sqrt{ 21 } }{ 2 },& x_2=\frac{ -1+\sqrt{ 21 } }{ 2 }\end{array} \\&\begin{array} { l }x_1\approx-2.79129,& x_2\approx1.79129\end{array}\end{align*}$

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