Example 1
Completion requirements
| Example 1 |
Solve the inequality \(x^2 + 2x + 1 \le 0\). Plot the solution set on a number line.
Determine the zeros of the corresponding function, \(f(x) = x^2 + 2x + 1\), to determine possible
solution intervals.
\(\begin{align}
x^2 + 2x + 1 &= 0 \\
\left( {x + 1} \right)^2 &= 0
\end{align}\)
\(\begin{align}
x + 1 &= 0 \\
x &= - 1 \\
\end{align}\)
There is only one unique zero, so there are two intervals that may be part of the solution.
Select a test point from each interval, and substitute these values into the original inequality to see which interval(s) are part of the solution.
A test point can be any value that is not on the boundary of the interval of interest. Try to use simple values whenever possible.The test point \(–5\) does not satisfy the inequality, so no values in the blue region are part of the solution set.
Test the interval \(x \ge -1\) (the green interval). The test point \(x = 0\) is used.
The test point \(0\) does not satisfy the inequality, so no values in the green region are part of the solution set.
Neither of the intervals satisfy the inequality, so \(x = -1\) is the solution to the inequality. Plot the solution on a number line.
solution intervals.
\(\begin{align}
x^2 + 2x + 1 &= 0 \\
\left( {x + 1} \right)^2 &= 0
\end{align}\)
\(\begin{align}
x + 1 &= 0 \\
x &= - 1 \\
\end{align}\)
There is only one unique zero, so there are two intervals that may be part of the solution.
Select a test point from each interval, and substitute these values into the original inequality to see which interval(s) are part of the solution.
Test the interval \(x \le -1\) (the blue interval). The test point \(x = -5\) is used.
| Left Side | Right Side |
|---|---|
|
\(\begin{array}{r} x^2 + 2x + 1 \\ \left( { - 5} \right)^2 + 2\left( { - 5} \right) + 1 \\ 16 \end{array}\) |
\(0\) |
| LS \(\gt\) RS\(\hspace{30pt}\) | |
A test point can be any value that is not on the boundary of the interval of interest. Try to use simple values whenever possible.
Test the interval \(x \ge -1\) (the green interval). The test point \(x = 0\) is used.
| Left Side | Right Side |
|---|---|
|
\(\begin{array}{r} x^2 + 2x + 1 \\ \left( 0 \right)^2 + 2\left( 0 \right) + 1 \\ 1 \end{array}\) |
\(0\) |
| LS \(\gt\) RS \(\hspace{30pt}\) | |
The test point \(0\) does not satisfy the inequality, so no values in the green region are part of the solution set.
Neither of the intervals satisfy the inequality, so \(x = -1\) is the solution to the inequality. Plot the solution on a number line.