When solving inequalities, you are trying to identify a set of numbers that will make the inequality true, and then represent the those intervals, usually an interval, on a number line.
You will apply the same strategies for solving quadratic equations when solving quadratic inequalities.
When solving inequalities, you are trying to identify a set of numbers that will make the inequality true, and then represent the those intervals, usually an interval, on a number line.
Your game plan for solving quadratic inequalities algebraically is to
The interactive below shows you how to determine the roots of related quadratic equations using an algebraic solution method. Click on the "begin" button to see the steps and the description of the steps used to identify those roots.
Now that you have found the roots in the three examples, you will need to examine those roots to determine which interval will satisfy the inequality. You can use a couple of different methods to make your determination.
You can pick a test point to substitute into the inequality based on the interval created from the roots, or you can create a sign chart based on the roots. In both methods, you are trying to find each interval that will satisfy the inequality.
Using the inequality from example 1, 3x2 ≥ 27, plot the roots on the x-axis. If you were looking at the solution graphically, you could simply find the interval of the graph that satisfies the inequality. Algebraically, you will need to look at the x-axis as a number line. It will be broken into intervals by the roots of our quadratic inequality.
Click on the image below to animate.
For the inequality, 3x2 ≥ 27, there are three distinct intervals on the number line.
Left Interval | Middle Interval | Right Interval |
(-∞, -3] | [-3, 3] | [3, ∞) |
x ≤ -3 | -3 ≤ x ≤ 3 | x ≥ 3 |
To test each interval, select a point on the number line that lies in the interval. Substitute the value for x into the inequality, and see if it makes the inequality true. Click on the different sections of the number line to see how sample test points are used to determine the solution.
Another strategy is to create a sign chart. Basically, you will use the same process as when you used test points. The difference is that you will use the sign of both the factors and their products to compare with the inequality. For this, you will need to plot the roots of the inequality on a number line. Be sure to check to see if the inequality sign is strictly a greater or less than symbol. If this is the case, the roots will not be included.
For this method, let’s use example 2 from above. You have already determined the roots of the quadratic inequality, x2 + 5x − 6 < 0, to be (-6, 0) and (1, 0). Rewriting the inequality in factored form, (x − 6)(x + 1) < 0, allows us to make a chart to find where the product of the two factors is less than 0.
To create a sign chart, select a point on the number line that lies in the test interval. Substitute the value for x into the inequality, and see what sign each root is. Click on the different sections of the number line to see how sample test points are used to determine the solution.
What are some important things to consider when selecting test values for inequalities?
Interactive popup. Assistance may be required. You would want to select values that are easy to add, subtract, multiply or divide. Often -1, 1 or 0 are good examples.
How is solving quadratic inequalities algebraically different from solving quadratic equations algebraically?
Interactive popup. Assistance may be required. The processes are exactly the same. The only difference is that after you find the roots of the quadratic equation, you must determine which interval created by the roots will satisfy the inequality.
Consider for a moment the three different algebraic approaches used to solve quadratic inequality above. In each one, the goal was to find the roots of the quadratic inequality. Why might you use the quadratic formula instead of factoring or inverse operations?
Will you be able to solve any quadratic inequality algebraically?
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Think back to example 2. Rewrite the inequality in standard form for a quadratic equation, and then determine the roots of the related quadratic equation.
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The roots of the related quadratic equation are (−2, 0) and (6, 0).
Use these roots as boundary points for test intervals.
| Interval (-∞, -2] | Interval [-2, 6] | Interval [6, ∞) |
| Inequality x ≤ -2 | Inequality -2 ≤ x ≤ 6 | Inequality x ≥ 6 |
| Example test point: -3 | Example test point: 0 | Example test point: 7 |
x2 ≤ 12 + 4x |
x2 ≤ 12 + 4x |
x2 ≤ 12 + 4x |
| False | True | False |
The solution to x 2 ≤ 12 + 4x is -2 ≤ x ≤ 6 .
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Think back to example 3. Rewrite the inequality in standard form, and then use the quadratic formula to determine the roots of the related quadratic equation.
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The roots of the related quadratic equation are (1, 0) and ( 3 over 2 - 3 2 , 0 ) .
| Interval (-∞, - 3 over 2 3 2 ] | Interval [- 3 over 2 3 2 , 1] | Interval [1, ∞) |
| Inequality x < - 3 over 2 3 2 | Inequality - 3 over 2 3 2 < x < 1 | Inequality x > 1 |
| Example test point: -2 | Example test point: 0 | Example test point: 2 |
3 > 2x2 + x |
3 > 2x2 + x |
3 > 2x2 + x |
| False | True | False |
The solution to 3 > 2x2 + x is - 3 over 2 3 2 < x < 1 .
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You can look at this problem several ways. You could approach this problem using inverse operations. Just remember to consider both positive and negative square roots. Look at example 1 if you need a jump start.
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The roots of the related quadratic equation are ( - 1 over 4 1 4 , 0 ) and ( 1 over 4 1 4 , 0 ) .
| Interval (-∞, - 1 over 1 1 4 ] | Interval [- 1 over 4 1 4 , 1 over 4 1 4 ] | Interval [ 1 over 4 1 4 , ∞) |
| Inequality x ≤ - 1 over 4 1 4 | Inequality - 1 over 4 1 4 ≤ x ≤ 1 over 4 1 4 | Inequality x ≥ 1 over 4 1 4 |
| Example test point: -1 | Example test point: 0 | Example test point: 1 |
1 ≤ 16x2 |
1 ≤ 16x2 |
1 ≤ 16x2 |
| True | False | True |
The solution to 1 ≤ 16x2 is x ≤ - 1 over 4 1 4 or x ≥ 1 over 4 1 4 .