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Completion requirements
Solve the inequality \(5x + 3 \gt 2x^2\) using sign analysis. Plot the solution set on a number line.
You may have noticed that the three methods shown for solving a quadratic inequality in one variable follow a general pattern:
Once you have solved an inequality, you can use all or part of another method to verify the solution. For example, after solving an inequality using sign analysis, you can choose a test point from each interval to verify the solution.
\(\begin{align}
5x + 3 &\gt 2x^2 \\
0 &\gt 2x^2 - 5x - 3 \\
0 &\gt 2x^2 - 6x + x - 3 \\
0 &\gt 2x\left( {x - 3} \right) + \left( {x - 3} \right) \\
0 &\gt \left( {x - 3} \right)\left( {2x + 1} \right) \\
\end{align}\)
The inequality \(0 \gt 2x^2 - 5x - 3\) is true for \(-\frac{1}{2} \lt x \lt 3\), so the solution to \(5x + 3 \gt 2x^2\) is \(\left\{ {x \thinspace | - \frac{1}{2} \lt x \lt 3, x \in \thinspace {\rm{R}}} \right\}\).
5x + 3 &\gt 2x^2 \\
0 &\gt 2x^2 - 5x - 3 \\
0 &\gt 2x^2 - 6x + x - 3 \\
0 &\gt 2x\left( {x - 3} \right) + \left( {x - 3} \right) \\
0 &\gt \left( {x - 3} \right)\left( {2x + 1} \right) \\
\end{align}\)
The inequality \(0 \gt 2x^2 - 5x - 3\) is true for \(-\frac{1}{2} \lt x \lt 3\), so the solution to \(5x + 3 \gt 2x^2\) is \(\left\{ {x \thinspace | - \frac{1}{2} \lt x \lt 3, x \in \thinspace {\rm{R}}} \right\}\).
You may have noticed that the three methods shown for solving a quadratic inequality in one variable follow a general pattern:
- Rearrange the inequality such that one side is \(0\).
- Use the non-zero side of the inequality to determine intervals such that the non-zero side is positive or negative for the entirety of each interval.
- Determine whether the non-zero side is positive or negative on each interval.
- Combine the appropriate intervals to form the solution set.
Once you have solved an inequality, you can use all or part of another method to verify the solution. For example, after solving an inequality using sign analysis, you can choose a test point from each interval to verify the solution.