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Solve an expression involving the absolute value
Answer
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$y = x ^ { 2 } - \left ( | x | \right ) - 20$
$y = 0$
$x$Intercept
$\left ( - 5 , 0 \right )$, $\left ( 5 , 0 \right )$
$y$Intercept
$\left ( 0 , - 20 \right )$
$\begin{array} {l} x = - 5 \\ x = 5 \end{array}$
$ $ Solve a solution to $ x$
$x ^ { 2 } - \left ( | x | \right ) \color{#FF6800}{ - } \color{#FF6800}{ 20 } = 0$
$ $ Move the constant to the right side and change the sign $ $
$x ^ { 2 } - \left ( | x | \right ) = \color{#FF6800}{ 20 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \left ( | \color{#FF6800}{ x } | \right ) = \color{#FF6800}{ 20 }$
$ $ Divide the interval based on the value where the inside of the absolute value is 0 $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \right ) = \color{#FF6800}{ 20 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ 0 } \right ) \\ \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } = \color{#FF6800}{ 20 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ 0 } \right )$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \right ) = \color{#FF6800}{ 20 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ 0 } \right ) \\ \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ x } = \color{#FF6800}{ 20 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ 0 } \right )$
$ $ Find the solution $ $
$\left. \begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \end{array} \right\} \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ 0 } \right ) \\ \left. \begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \end{array} \right\} \left ( $ However (or only) $ \color{#FF6800}{ x } \geq \color{#FF6800}{ 0 } \right )$
$\left. \begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \end{array} \right\} \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ 0 } \right ) \\ \left. \begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \end{array} \right\} \left ( $ However (or only) $ \color{#FF6800}{ x } \geq \color{#FF6800}{ 0 } \right )$
$ $ Make sure if the value is within the interval $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 5 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 5 }$
$ $ Find the union of sets of each interval $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 5 } \end{array}$
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