# Solve for: |(x^2-3)/(x^2+x)| >= 1

## Expression: $|\frac{ {x}^{2}-3 }{ {x}^{2}+x }| \geq 1$

Determine the defined range

$\begin{array} { l }|\frac{ {x}^{2}-3 }{ {x}^{2}+x }| \geq 1,& \begin{array} { l }x≠0,& x≠-1\end{array}\end{array}$

Separate the inequality into $2$ possible cases

$\begin{array} { l }\begin{array} { l }\frac{ {x}^{2}-3 }{ {x}^{2}+x } \geq 1,& \frac{ {x}^{2}-3 }{ {x}^{2}+x } \geq 0\end{array},\\\begin{array} { l }-\frac{ {x}^{2}-3 }{ {x}^{2}+x } \geq 1,& \frac{ {x}^{2}-3 }{ {x}^{2}+x } < 0\end{array}\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\begin{array} { l }x \in \left\langle-\infty, -3\right] \cup \langle-1, 0\rangle,& \frac{ {x}^{2}-3 }{ {x}^{2}+x } \geq 0\end{array},\\\begin{array} { l }-\frac{ {x}^{2}-3 }{ {x}^{2}+x } \geq 1,& \frac{ {x}^{2}-3 }{ {x}^{2}+x } < 0\end{array}\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\begin{array} { l }x \in \left\langle-\infty, -3\right] \cup \langle-1, 0\rangle,& x \in \left\langle-\infty, -\sqrt{ 3 }\right] \cup \langle-1, 0\rangle \cup \left[ \sqrt{ 3 }, +\infty\right\rangle\end{array},\\\begin{array} { l }-\frac{ {x}^{2}-3 }{ {x}^{2}+x } \geq 1,& \frac{ {x}^{2}-3 }{ {x}^{2}+x } < 0\end{array}\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\begin{array} { l }x \in \left\langle-\infty, -3\right] \cup \langle-1, 0\rangle,& x \in \left\langle-\infty, -\sqrt{ 3 }\right] \cup \langle-1, 0\rangle \cup \left[ \sqrt{ 3 }, +\infty\right\rangle\end{array},\\\begin{array} { l }x \in \left[ -\frac{ 3 }{ 2 }, -1\right\rangle \cup \left\langle0, 1\right],& \frac{ {x}^{2}-3 }{ {x}^{2}+x } < 0\end{array}\end{array}$

Solve the inequality for $x$

$\begin{array} { l }\begin{array} { l }x \in \left\langle-\infty, -3\right] \cup \langle-1, 0\rangle,& x \in \left\langle-\infty, -\sqrt{ 3 }\right] \cup \langle-1, 0\rangle \cup \left[ \sqrt{ 3 }, +\infty\right\rangle\end{array},\\\begin{array} { l }x \in \left[ -\frac{ 3 }{ 2 }, -1\right\rangle \cup \left\langle0, 1\right],& x \in \langle-\sqrt{ 3 }, -1\rangle \cup \langle0, \sqrt{ 3 }\rangle\end{array}\end{array}$

Find the intersection

$\begin{array} { l }x \in \left\langle-\infty, -3\right] \cup \langle-1, 0\rangle,\\\begin{array} { l }x \in \left[ -\frac{ 3 }{ 2 }, -1\right\rangle \cup \left\langle0, 1\right],& x \in \langle-\sqrt{ 3 }, -1\rangle \cup \langle0, \sqrt{ 3 }\rangle\end{array}\end{array}$

Find the intersection

$\begin{array} { l }x \in \left\langle-\infty, -3\right] \cup \langle-1, 0\rangle,\\x \in \left[ -\frac{ 3 }{ 2 }, -1\right\rangle \cup \left\langle0, 1\right]\end{array}$

Find the union

$\begin{array} { l }x \in \left\langle-\infty, -3\right] \cup \left[ -\frac{ 3 }{ 2 }, -1\right\rangle \cup \langle-1, 0\rangle \cup \left\langle0, 1\right],& \begin{array} { l }x≠0,& x≠-1\end{array}\end{array}$

Find the intersection of the solution and the defined range

$x \in \left\langle-\infty, -3\right] \cup \left[ -\frac{ 3 }{ 2 }, -1\right\rangle \cup \langle-1, 0\rangle \cup \left\langle0, 1\right]$

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