Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Algebra 2 Lesson 1-6 Part 2 Absolute Value Inequalities Goals Goal • To write and solve inequalities involving absolute values. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems. Essential Question Big Idea: Absolute Value Inequalities • How can you solve an absolute value inequality? • Students will understand that an absolute value quantity is nonnegative. Vocabulary • None Absolute Value Inequalities • When an inequality contains an absolute-value expression, it can be written as a compound inequality. • The inequality |x| < 5 describes all real numbers whose distance from 0 is less than 5 units. • The solutions are all numbers between –5 and 5, so |x|< 5 can be rewritten as –5 < x < 5, or as x > –5 AND x < 5. “Less Than” - Conjunction The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is a conjunction: –3 < x < 3. Helpful Hint Think: Less thand inequalities involving < or ≤ symbols are conjunctions. “Less Than” Absolute Value Inequalities Example: Solve the inequality and graph the solutions. |x|– 3 < –1 Since 3 is subtracted from |x|, add 3 to both sides to undo the subtraction. |x|– 3 < –1 +3 +3 |x| < 2 x > –2 AND x < 2 2 units –2 –1 Write as a compound inequality. The solution set is {x: –2 < x < 2}. 2 units 0 1 2 Example: Solve the inequality and graph the solutions. |x – 1| ≤ 2 x – 1 ≥ –2 AND x – 1 ≤ 2 +1 +1 +1 +1 x ≥ –1 AND Write as a compound inequality. Solve each inequality. x≤3 Write as a compound inequality. The solution set is {x: –1 ≤ x ≤ 3}. –3 –2 –1 0 1 2 3 Procedure Solving “Less Than” Inequalities Solving Absolute Inequalities: 1. Isolate the absolute value. 2. Break it into 2 inequalities (and statement) – one positive and the other negative reversing the sign. 3. Solve both inequalities. 4. Check both answers. Helpful Hint Just as you do when solving absolute-value equations, you first isolate the absolute-value expression when solving absolute-value inequalities. Your Turn: Solve the inequality and graph the solutions. 2|x| ≤ 6 2|x| ≤ 6 2 2 |x| ≤ 3 Since x is multiplied by 2, divide both sides by 2 to undo the multiplication. Write as a compound inequality. The solution set is {x: –3 ≤ x ≤ 3}. x ≥ –3 AND x ≤ 3 3 units –3 –2 –1 3 units 0 1 2 3 Your Turn: Solve each inequality and graph the solutions. |x + 3|– 4.5 ≤ 7.5 |x + 3|– 4.5 ≤ 7.5 + 4.5 +4.5 Since 4.5 is subtracted from |x + 3|, add 4.5 to both sides to undo the subtraction. |x + 3| ≤ 12 x + 3 ≥ –12 AND x + 3 ≤ 12 –3 –3 –3 –3 Write as a compound inequality. The solution set is {x: –15 ≤ x ≤ 9}. x ≥ –15 AND x ≤ 9 –20 –15 –10 –5 0 5 10 15 “Greater Than” Absolute Value Inequalities • The inequality |x| > 5 describes all real numbers whose distance from 0 is greater than 5 units. • The solutions are all numbers less than –5 or greater than 5. • The inequality |x| > 5 can be rewritten as the compound inequality x < –5 OR x > 5. “Greater Than” - Disjunction The solutions of |x| > 3 are the points that are more than 3 units from zero. The solution is a disjunction: x < –3 or x > 3. Helpful Hint Think: Greator inequalities involving > or ≥ symbols are disjunctions. “Greater Than” Absolute Value Inequalities Example: Solve the inequality and graph the solutions. |x| + 14 ≥ 19 |x| + 14 ≥ 19 – 14 –14 |x| ≥ 5 x ≤ –5 OR x ≥ 5 Since 14 is added to |x|, subtract 14 from both sides to undo the addition. Write as a compound inequality. The solution set is {x: x ≤ –5 OR x ≥ 5}. 5 units –10 –8 –6 –4 –2 5 units 0 2 4 6 8 10 Example: Solve the inequality and graph the solutions. 3 + |x + 2| > 5 Since 3 is added to |x + 2|, subtract 3 from both sides to undo the addition. 3 + |x + 2| > 5 –3 –3 |x + 2| > 2 x + 2 < –2 OR x + 2 > 2 –2 –2 –2 –2 Write as a compound inequality. Solve each inequality. Write as a compound inequality. x < –4 OR x > 0 –10 –8 –6 –4 –2 The solution set is {x: x < –4 or x > 0}. 0 2 4 6 8 10 Procedure Solving “Greater Than” Inequalities Solving Absolute Inequalities: 1. Isolate the absolute value. 2. Break it into 2 inequalities (or statement) – one positive and the other negative reversing the sign. 3. Solve both inequalities. 4. Check both answers. Your Turn: Solve each inequality and graph the solutions. |x| + 10 ≥ 12 |x| + 10 ≥ 12 – 10 –10 |x| Since 10 is added to |x|, subtract 10 from both sides to undo the addition. ≥ 2 x ≤ –2 OR x ≥ 2 Write as a compound inequality. The solution set is {x: x ≤ –2 or x ≥ 2}. 2 units 2 units –5 –4 –3 –2 –1 0 1 2 3 4 5 Your Turn: Solve the inequality and graph the solutions. Since is added to |x + 2 |, subtract from both sides to undo the addition. Write as a compound inequality. Solve each inequality. OR x ≤ –6 x≥1 –7 –6 –5 –4 –3 –2 –1 Write as a compound inequality. The solution set is {x: x ≤ –6 or x ≥ 1} 0 1 2 3 Example: Application A pediatrician recommends that a baby’s bath water be 95°F, but it is acceptable for the temperature to vary from this amount by as much as 3°F. Write and solve an absolute-value inequality to find the range of acceptable temperatures. Graph the solutions. Let t represent the actual water temperature. The difference between t and the ideal temperature is at most 3°F. t – 95 ≤ 3 Example: Continued t – 95 ≤ 3 |t – 95| ≤ 3 t – 95 ≥ –3 AND t – 95 ≤ 3 +95 +95 +95 +95 t ≥ 92 AND t 90 Solve the two inequalities. ≤ 98 92 94 96 98 100 The range of acceptable temperature is 92 ≤ t ≤ 98. Your Turn: A dry-chemical fire extinguisher should be pressurized to 125 psi, but it is acceptable for the pressure to differ from this value by at most 75 psi. Write and solve an absolute-value inequality to find the range of acceptable pressures. Graph the solution. Let p represent the desired pressure. The difference between p and the ideal pressure is at most 75 psi. p – 125 ≤ 75 Your Turn: Continued p – 125 ≤ 75 |p – 125| ≤ 75 p – 125 ≥ –75 AND p – 125 ≤ 75 +125 +125 +125 +125 p ≥ 50 AND p 25 50 Solve the two inequalities. ≤ 200 75 100 125 150 175 200 The range of pressure is 50 ≤ p ≤ 200. 225 Solutions to Absolute Value Inequalities • When solving an absolute-value inequality, you may get a statement that is true for all values of the variable. • In this case, all real numbers are solutions of the original inequality. • If you get a false statement when solving an absolutevalue inequality, the original inequality has no solutions. Its solution set is ø. Example: Solve the inequality. |x + 4|– 5 > – 8 |x + 4|– 5 > – 8 +5 +5 |x + 4| > –3 Add 5 to both sides. Absolute-value expressions are always nonnegative. Therefore, the statement is true for all real numbers. The solution set is all real numbers. Example: Solve the inequality. |x – 2| + 9 < 7 |x – 2| + 9 < 7 –9 –9 |x – 2| < –2 Subtract 9 from both sides. Absolute-value expressions are always nonnegative. Therefore, the statement is false for all values of x. The inequality has no solutions. The solution set is ø. Remember! An absolute value represents a distance, and distance cannot be less than 0. Your Turn: Solve the inequality. |x| – 9 ≥ –11 |x| – 9 ≥ –11 +9 ≥ +9 |x| ≥ –2 Add 9 to both sides. Absolute-value expressions are always nonnegative. Therefore, the statement is true for all real numbers. The solution set is all real numbers. Your Turn: Solve the inequality. 4|x – 3.5| ≤ –8 4|x – 3.5| ≤ –8 4 4 |x – 3.5| ≤ –2 Divide both sides by 4. Absolute-value expressions are always nonnegative. Therefore, the statement is false for all values of x. The inequality has no solutions. The solution set is ø. SUMMARY Absolute Value Inequalities (< , ) Inequalities of the Form < or Involving Absolute Value If a is a positive real number and if u is an algebraic expression, then |u| < a is equivalent to u < a and u > -a |u| a is equivalent to u a and u > -a Note: If a = 0, |u| < 0 has no real solution, |u| 0 is equivalent to u = 0. If a < 0, the inequality has no real solution. Absolute Value Inequalities (> , ) Inequalities of the Form > or Involving Absolute Value If a is a positive real number and u is an algebraic expression, then |u| > a is equivalent to u < – a or u > a |u| a is equivalent to u – a or u a. Procedure Solving Absolute Inequalities: 1. Isolate the absolute value. 2. Break it into 2 inequalites (“< , ≤ and statement” – “> , ≥ or statement”) – one positive and the other negative reversing the sign. 3. Solve both inequalities. 4. Check both answers. Solving an Absolute-Value Equation Recall that |xx is | isthe thedistance distancebetween betweenxxand and0.0.IfIf x | x | 8,8,then then any number between 8 and 8 is a solution of the inequality. 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 You can use the following properties to solve absolute-value inequalities and equations. 7 8 SOLVING ABSOLUTE-VALUE INEQUALITIES SOLVING ABSOLUTE-VALUE INEQUALITIES | ax b | c means ax b c and a x b c. a number, means valueais | a xWhen b | an c absolute x less b than c and a x the b c. inequalities are connected by and. When an absolute value is greater than a number, the inequalities are connected by or. | ax b | c means ax b c or a x b c. | ax b | c means ax b c or a x b c. Solving an Absolute-Value Inequality Solve | x 4 | < 3 x 4 IS POSITIVE x 4 IS NEGATIVE |x4|3 |x4|3 x 4 3 x 4 3 x7 x1 Reverse inequality symbol. The solution is all real numbers greater than 1 and less than 7. This can be written as 1 x 7. –10 –8 –6 –4 –2 0 2 4 6 8 10 Solving an Absolute-Value Inequality Solve 1 | 3 6 and graph 2x + 1| 2x IS POSITIVE 2x +the 1 ISsolution. NEGATIVE | 2x 1 | 3 6 | 2x 1 | 3 6 2x + 1 IS POSITIVE | 2x|2x1 | 13| 69 2x + 1 IS NEGATIVE 1 | 69 | 2x|2x1 | 3 2x 1 9 | 2x 1 | 9 2x 10 2x 8 2x 1 9 2x 1 +9 x4 x 5 2x 10 2x 8 The solution is all real numbers greater than or equal x4 x 5 to 4 or less than or equal to 5. This can be written as the compound inequality x 5 or x 4. Reverse 2x11| +9 | 2x 9 inequality symbol. 6 5 4 3 2 1 0 1 2 3 4 5 6 MORE PROBLEMS FOR YOU Your Turn: Solve the compound inequality. Then graph the solution set. |p – 2| ≤ –6 |p – 2| ≤ –6 and p – 2 ≥ 6 p ≤ –4 and p ≥ 8 Multiply both sides by –2, and reverse the inequality symbol. Rewrite the absolute value as a conjunction. Add 2 to both sides of each inequality. Because no real number satisfies both p ≤ –4 and p ≥ 8, there is no solution. The solution set is ø. Your Turn: Solve the compound inequality. Then graph the solution set. |x – 5| ≤ 8 Multiply both sides by 2. x – 5 ≤ 8 and x – 5 ≥ –8 Rewrite the absolute value as a conjunction. x ≤ 13 and x ≥ –3 Add 5 to both sides of each inequality. Continued The solution set is {x|–3 ≤ x ≤ 13}. –10 –5 0 5 10 15 20 25 Your Turn: Solve the compound inequality. Then graph the solution set. |2x +7| ≤ 3 Multiply both sides by 3. 2x + 7 ≤ 3 and 2x + 7 ≥ –3 Rewrite the absolute value as a conjunction. 2x ≤ –4 and 2x ≥ –10 Subtract 7 from both sides of each inequality. x ≤ –2 and x ≥ –5 Divide both sides of each inequality by 2. Continued The solution set is {x|–5 ≤ x ≤ 2}. –6 –5 –3 –2 –1 0 1 2 3 4 Your Turn: Solve the compound inequality. Then graph the solution set. –2|x +5| > 10 |x + 5| < –5 x + 5 < –5 and x + 5 > 5 x < –10 and x > 0 Divide both sides by –2, and reverse the inequality symbol. Rewrite the absolute value as a conjunction. Subtract 5 from both sides of each inequality. Because no real number satisfies both x < –10 and x > 0, there is no solution. The solution set is ø. Your Turn: Solve: 1. |3k| + 11 > 8 2. –2|u + 7| ≥ 16 3. |1 – 2x| > 7 All Real Numbers No Solution x < –3 or x > 4 Assignment • Section 1-6 Part 2, Pg 49 – 52; #1 – 5 all, 6 – 52 even