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Completion requirements
Solve \(5x + 13 = \left|5(x + 1)^2 - 2\right|\) algebraically.
Case 1: \(5(x + 1)^2 - 2 \ge 0\)
\(\begin{align}
5x + 13 &= \left| {5\left( {x + 1} \right)^2 - 2} \right| \\
5x + 13& = 5\left( {x + 1} \right)^2 - 2 \\
5x + 13 &= 5\left( {x^2 + 2x + 1} \right) - 2 \\
5x + 13 &= 5x^2 + 10x + 5 - 2 \\
0 &= 5x^2 + 5x - 10 \\
0 &= x^2 + x - 2 \\
0 &= \left( {x + 2} \right)\left( {x - 1} \right) \\
\end{align}\)
These potential solutions can be verified graphically, as shown, or algebraic verification is also appropriate.
Case 2: \(5(x + 1)^2 - 2 \lt 0\)
\(\begin{align}
5x + 13 &= \left| {5\left( {x + 1} \right)^2 - 2} \right| \\
5x + 13 &= - \left( {5\left( {x + 1} \right)^2 - 2} \right) \\
5x + 13 &= - 5\left( {x^2 + 2x + 1} \right) + 2 \\
5x + 13 &= - 5x^2 - 10x - 5 + 2 \\
5x^2 + 15x + 16 &= 0 \\
\end{align}\)
The discriminant is negative, so there are no real solutions to this case.
The solutions to the equation are \(-2\) and \(1\).
\(\begin{align}
5x + 13 &= \left| {5\left( {x + 1} \right)^2 - 2} \right| \\
5x + 13& = 5\left( {x + 1} \right)^2 - 2 \\
5x + 13 &= 5\left( {x^2 + 2x + 1} \right) - 2 \\
5x + 13 &= 5x^2 + 10x + 5 - 2 \\
0 &= 5x^2 + 5x - 10 \\
0 &= x^2 + x - 2 \\
0 &= \left( {x + 2} \right)\left( {x - 1} \right) \\
\end{align}\)
\(\begin{align}
0 &= x + 2 \\
- 2 &= x \\
\end{align}\)
0 &= x + 2 \\
- 2 &= x \\
\end{align}\)
\(\begin{align}
0 &= x - 1 \\
1 &= x \\
\end{align}\)
0 &= x - 1 \\
1 &= x \\
\end{align}\)
These potential solutions can be verified graphically, as shown, or algebraic verification is also appropriate.
Case 2: \(5(x + 1)^2 - 2 \lt 0\)
\(\begin{align}
5x + 13 &= \left| {5\left( {x + 1} \right)^2 - 2} \right| \\
5x + 13 &= - \left( {5\left( {x + 1} \right)^2 - 2} \right) \\
5x + 13 &= - 5\left( {x^2 + 2x + 1} \right) + 2 \\
5x + 13 &= - 5x^2 - 10x - 5 + 2 \\
5x^2 + 15x + 16 &= 0 \\
\end{align}\)
\[\begin{align}
x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
&= \frac{{ - 15 \pm \sqrt {\left( {15} \right)^2 - 4\left( 5 \right)\left( {16} \right)} }}{{2\left( 5 \right)}} \\
&= \frac{{ - 15 \pm \sqrt { - 95} }}{{10}} \\
\end{align}\]
x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
&= \frac{{ - 15 \pm \sqrt {\left( {15} \right)^2 - 4\left( 5 \right)\left( {16} \right)} }}{{2\left( 5 \right)}} \\
&= \frac{{ - 15 \pm \sqrt { - 95} }}{{10}} \\
\end{align}\]
The discriminant is negative, so there are no real solutions to this case.
The solutions to the equation are \(-2\) and \(1\).