How to Solve Quadratic Inequalities
Quadratic equations are polynomial equations in which the leading term variable is squared and, to solve them, the process of factoring needs to be utilized. Factoring is the breakdown of polynomial equations into simpler equations, which, when multiplied together, give the same result. The inequality aspect means that the solution is not equal to only one number. The factoring aspect of quadratics can create more than one solution that can all hold true or be any one of of several. Factoring is the key process to solving quadratic inequalities.
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Instructions
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Inequalities With Greater Than or Less Than
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1
Write the inequality on paper.
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2
Simplify the inequality, if possible. It is not possible here.
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3
Get all terms on one side of the inequality and 0 on the other. Subtract 10 from both sides.
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4
Factor the terms. The factors are x + 5 and x - 2.
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5
Decide the positive and negative combinations required to satisfy the inequality. A positive is required, so both factors need to be positive. This results in two possible combinations.
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6
Write each factor as an inequality equal to 0 with the signs determined in Step 5. These are the critical values.
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7
Solve each inequality for x.
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8
Determine which critical value satisfy the inequality and this is the answer. Sometimes both sets of critical values can be part of the solution. Both sets are possible.
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9
Simplify the answer, if possible. X is greater than 2 or less than -5.
Inequalities With Greater Than or Equal To or Less Than or Equal To
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10
Write the inequality on paper.
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11
Simplify the inequality, if possible. Distribute the -x and divide both sides by -2, which means the inequality sign is reversed.
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12
Get all terms on one side of the inequality and 0 on the other. Add 12 to both sides
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13
Factor the terms. The factors are x - 3 and x - 4.
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14
Decide the positive and negative combinations required to satisfy the inequality. A negative is required, so one factor needs to be negative and the other a positive. This results in two possible combinations.
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15
Write each factor as an inequality equal to 0 with the signs determined in Step 5. These are the critical values.
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16
Solve each inequality for x.
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17
Determine which critical value(s) satisfy the inequality and this is the answer. Sometimes both sets of critical values can be part of the solution. The first set is not possible, so it cannot be a solution. The second set works and that is the answer.
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18
Simplify the answer, if possible. X is greater than or equal to 3 AND less than or equal to 4.
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1
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Comments
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1350ng
Oct 13, 2010
Sorry for the mistake. The equation is x^2 + 3x - 10 > 0. Two real roots -5 and 2. Test-point x = 0 gives: -10 > 0. It is not true, then the origin is not located on the TRUE segment. The solution set are the 2 intervals (-infinity, -5) (2, +infinity)
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1350ng
Oct 13, 2010
The approach to determine "which critical values satisfy the inequality" may be confusing and consumes too much time. During tests/exams time is limited. The best way to quickly obtain the solution set is to use the number-line method and the test-point approach. In the first example, x^2 + 3x - 10 < 0, plot the 2 real roots -5 and 2 on the number line. Use as test-point the origin x = 0. It gets: -10 < 0. It is not true, then the origin is not on the true segment. The solution set are the two intervals: (-infinity, -5) and (2, +infinity). The two end-points -5 and 2 are not included in the solution set.
