Example 1
Completion requirements
| Example 1 |
Multimedia |
Solve \(\left| x^2 - 2x - 3\right| = 3\) algebraically.
Solving quadratic inequalities will be covered in Unit 7. In this unit, verification by substitution or by graphing will be used to check for extraneous roots.
If you come up with a way to determine the \(x\)-values that satisfy a quadratic inequality, such as \(x^2 - 2x - 3 \ge 0\), you are free to use it in your solution.
The values \(1 + \sqrt {7}\) and \(1 - \sqrt {7}\) may be part of the solution. Verification is required to ensure they are not extraneous.
Verify for \(x = 1 + \sqrt {7}\).
Verify for \(x = 1 - \sqrt {7}\).
The values \(1 + \sqrt{7}\) and \(1 - \sqrt{7}\) are both solutions to the equation.
Case 2: \(x^2 - 2x - 3 \lt 0\)
In this case, \(x^2 - 2x - 3 \lt 0\), so it becomes negative when the absolute value symbol is dropped.
\(\begin{align}
\left| {x^2 - 2x - 3} \right| &= 3 \\
- \left( {x^2 - 2x - 3} \right) &= 3 \\
- x^2 + 2x + 3 &= 3 \\
0 &= x^2 - 2x \\
0 &= x\left( {x - 2} \right) \\
\end{align}\)
The values \(0\) and \(2\) may be part of the solution. Verification is required to ensure they are not extraneous.
The values \(0\) and \(2\) are both solutions to the equation.
The solutions to \(\left|x^2 - 2x - 3\right| = 3\) are \(1 - \sqrt{7},\thinspace 0,\thinspace 2,\) and \(1 + \sqrt{7}\).
Case 1: \(x^2 - 2x - 3 \ge 0\)
In this case, \(x^2 - 2x - 3 \ge 0\), so the absolute value symbol can be dropped without changing the expression.
\(\begin{align}
\left| {x^2 - 2x - 3} \right| &= 3 \\
x^2 - 2x - 3 &= 3 \\
x^2 - 2x - 6 &= 0 \\
\end{align}\)
In this case, \(x^2 - 2x - 3 \ge 0\), so the absolute value symbol can be dropped without changing the expression.
\(\begin{align}
\left| {x^2 - 2x - 3} \right| &= 3 \\
x^2 - 2x - 3 &= 3 \\
x^2 - 2x - 6 &= 0 \\
\end{align}\)
Solving quadratic inequalities will be covered in Unit 7. In this unit, verification by substitution or by graphing will be used to check for extraneous roots.
If you come up with a way to determine the \(x\)-values that satisfy a quadratic inequality, such as \(x^2 - 2x - 3 \ge 0\), you are free to use it in your solution.
\[\begin{align}
x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
&= \frac{{ - \left( { - 2} \right) \pm \sqrt {\left( { - 2} \right)^2 - 4\left( 1 \right)\left( { - 6} \right)} }}{{2\left( 1 \right)}} \\
&= \frac{{2 \pm \sqrt {28} }}{2} \\
&= \frac{{2 \pm 2\sqrt 7 }}{2} \\
&= 1 \pm \sqrt 7
\end{align}\]
x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
&= \frac{{ - \left( { - 2} \right) \pm \sqrt {\left( { - 2} \right)^2 - 4\left( 1 \right)\left( { - 6} \right)} }}{{2\left( 1 \right)}} \\
&= \frac{{2 \pm \sqrt {28} }}{2} \\
&= \frac{{2 \pm 2\sqrt 7 }}{2} \\
&= 1 \pm \sqrt 7
\end{align}\]
The values \(1 + \sqrt {7}\) and \(1 - \sqrt {7}\) may be part of the solution. Verification is required to ensure they are not extraneous.
Verify for \(x = 1 + \sqrt {7}\).
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} \left| {x^2 - 2x - 3} \right| \\ \left| {\left( {1 + \sqrt 7 } \right)^2 - 2\left( {1 + \sqrt 7 } \right) - 3} \right| \\ \left| {1 + 2\sqrt 7 + 7 - 2 - 2\sqrt 7 - 3} \right| \\ \left| 3 \right| \\ 3 \end{array}\) |
\(3\) |
| LS = RS\(\hspace{30pt}\) | |
Verify for \(x = 1 - \sqrt {7}\).
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} \left| {x^2 - 2x - 3} \right| \\ \left| {\left( {1 - \sqrt 7 } \right)^2 - 2\left( {1 - \sqrt 7 } \right) - 3} \right| \\ \left| {1 - 2\sqrt 7 + 7 - 2 + 2\sqrt 7 - 3} \right| \\ \left| 3 \right| \\ 3 \end{array}\) |
\(3\) |
| LS = RS \(\hspace{30pt}\) | |
The values \(1 + \sqrt{7}\) and \(1 - \sqrt{7}\) are both solutions to the equation.
Case 2: \(x^2 - 2x - 3 \lt 0\)
In this case, \(x^2 - 2x - 3 \lt 0\), so it becomes negative when the absolute value symbol is dropped.
\(\begin{align}
\left| {x^2 - 2x - 3} \right| &= 3 \\
- \left( {x^2 - 2x - 3} \right) &= 3 \\
- x^2 + 2x + 3 &= 3 \\
0 &= x^2 - 2x \\
0 &= x\left( {x - 2} \right) \\
\end{align}\)
\(0 = x\)
\(\begin{align}
0 &= x - 2 \\
2 &= x \\
\end{align}\)
0 &= x - 2 \\
2 &= x \\
\end{align}\)
Verify for \(x = 0\).
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} \left| {x^2 - 2x - 3} \right| \\ \left| {\left( 0 \right)^2 - 2\left( 0 \right) - 3} \right| \\ \left| { - 3} \right| \\ 3 \end{array}\) |
\(3\) |
| LS = RS \(\hspace{30pt}\) | |
Verify for \(x = 2\).
| Left Side | Right Side |
|---|---|
| \(\begin{array}{r} \left| {x^2 - 2x - 3} \right| \\ \left| {\left( 2 \right)^2 - 2\left( 2 \right) - 3} \right| \\ \left| { - 3} \right| \\ 3 \end{array}\) |
\(3\) |
| LS = RS \(\hspace{30pt}\) | |
The values \(0\) and \(2\) are both solutions to the equation.
The solutions to \(\left|x^2 - 2x - 3\right| = 3\) are \(1 - \sqrt{7},\thinspace 0,\thinspace 2,\) and \(1 + \sqrt{7}\).