College Algebra - Sections 1.4-1.6 Dr. Doug Ensley Shippensburg - PowerPoint PPT Presentation

College Algebra - Sections 1.4-1.6 Dr. Doug Ensley Shippensburg University April 20, 2015

Radicals The basic strategy for solving equations including radicals (such as √ , √ , √ , etc.: 3 4 1. Isolate the radical part of the equation. If there is more than one, isolate one of them. 2. Raise both sides to the appropriate power to eliminate the radical. Throughout this process, resist the urge to make shit up.

Radicals The basic strategy for solving equations including radicals (such as √ , √ , √ , etc.: 3 4 1. Isolate the radical part of the equation. If there is more than one, isolate one of them. 2. Raise both sides to the appropriate power to eliminate the radical. Throughout this process, resist the urge to make shit up. Example. Solve 3 √ 2 x + 1 − 7 = 8

Radicals The basic strategy for solving equations including radicals (such as √ , √ , √ , etc.: 3 4 1. Isolate the radical part of the equation. If there is more than one, isolate one of them. 2. Raise both sides to the appropriate power to eliminate the radical. Throughout this process, resist the urge to make shit up. Example. Solve 3 √ 2 x + 1 − 7 = 8 ◮ Solution. Add 7 to both sides: 3 √ 2 x + 1 = 15

Radicals The basic strategy for solving equations including radicals (such as √ , √ , √ , etc.: 3 4 1. Isolate the radical part of the equation. If there is more than one, isolate one of them. 2. Raise both sides to the appropriate power to eliminate the radical. Throughout this process, resist the urge to make shit up. Example. Solve 3 √ 2 x + 1 − 7 = 8 ◮ Solution. Add 7 to both sides: 3 √ 2 x + 1 = 15 ◮ Divide both sides by 3: √ 2 x + 1 = 5. ( Isolates the radical. )

Radicals The basic strategy for solving equations including radicals (such as √ , √ , √ , etc.: 3 4 1. Isolate the radical part of the equation. If there is more than one, isolate one of them. 2. Raise both sides to the appropriate power to eliminate the radical. Throughout this process, resist the urge to make shit up. Example. Solve 3 √ 2 x + 1 − 7 = 8 ◮ Solution. Add 7 to both sides: 3 √ 2 x + 1 = 15 ◮ Divide both sides by 3: √ 2 x + 1 = 5. ( Isolates the radical. ) ◮ Square both sides: 2 x + 1 = 25.

Radicals The basic strategy for solving equations including radicals (such as √ , √ , √ , etc.: 3 4 1. Isolate the radical part of the equation. If there is more than one, isolate one of them. 2. Raise both sides to the appropriate power to eliminate the radical. Throughout this process, resist the urge to make shit up. Example. Solve 3 √ 2 x + 1 − 7 = 8 ◮ Solution. Add 7 to both sides: 3 √ 2 x + 1 = 15 ◮ Divide both sides by 3: √ 2 x + 1 = 5. ( Isolates the radical. ) ◮ Square both sides: 2 x + 1 = 25. ◮ Continue until you’ve solved for x . Answer: x = 12

Radicals The basic strategy for solving equations including radicals (such as √ , √ , √ , etc.: 3 4 1. Isolate the radical part of the equation. If there is more than one, isolate one of them. 2. Raise both sides to the appropriate power to eliminate the radical. Throughout this process, resist the urge to make shit up. Example. Solve 3 √ 2 x + 1 − 7 = 8 ◮ Solution. Add 7 to both sides: 3 √ 2 x + 1 = 15 ◮ Divide both sides by 3: √ 2 x + 1 = 5. ( Isolates the radical. ) ◮ Square both sides: 2 x + 1 = 25. ◮ Continue until you’ve solved for x . Answer: x = 12 ◮ CHECK: 3 √ 2 · 12 + 1 − 7 = 3 √ 25 − 7 = 3 · 5 − 7 = 8

Examples Solve each of the following for x : 1. √ 2 x − 7 = 5 √ x 2 + 5 x = 9 2. 3 + 3. 4 + √ 5 x − 6 = x √ x 2 − x − 6 = x + 3 4.

More Examples Solve each of the following for x : √ x 2 + 4 x = 2 5 1. 2. (5 x + 6) 1 / 2 = 9 3. (3 x + 3) 1 / 3 = − 3 4. √ 3 x − 5 − √ x + 7 = 2

Quadratics with Substitution Sometimes we can transform complex equations into ones we know how to solve.

Quadratics with Substitution Sometimes we can transform complex equations into ones we know how to solve. Example. Solve the equation: x 4 − x 2 − 6 = 0

Quadratics with Substitution Sometimes we can transform complex equations into ones we know how to solve. Example. Solve the equation: x 4 − x 2 − 6 = 0 Solution. Substitute the variable u in place of x 2 . Note that since u = x 2 , it follows that u 2 = x 4 .

Quadratics with Substitution Sometimes we can transform complex equations into ones we know how to solve. Example. Solve the equation: x 4 − x 2 − 6 = 0 Solution. Substitute the variable u in place of x 2 . Note that since u = x 2 , it follows that u 2 = x 4 . ◮ The equation x 4 − x 2 − 6 = 0 becomes u 2 − u − 6 = 0.

Quadratics with Substitution Sometimes we can transform complex equations into ones we know how to solve. Example. Solve the equation: x 4 − x 2 − 6 = 0 Solution. Substitute the variable u in place of x 2 . Note that since u = x 2 , it follows that u 2 = x 4 . ◮ The equation x 4 − x 2 − 6 = 0 becomes u 2 − u − 6 = 0. ◮ The equation u 2 − u − 6 = 0 can be solved by factoring as ( u − 3)( u + 2) = 0 to conclude that either u = 3 or u = − 2.

Quadratics with Substitution Sometimes we can transform complex equations into ones we know how to solve. Example. Solve the equation: x 4 − x 2 − 6 = 0 Solution. Substitute the variable u in place of x 2 . Note that since u = x 2 , it follows that u 2 = x 4 . ◮ The equation x 4 − x 2 − 6 = 0 becomes u 2 − u − 6 = 0. ◮ The equation u 2 − u − 6 = 0 can be solved by factoring as ( u − 3)( u + 2) = 0 to conclude that either u = 3 or u = − 2. ◮ In terms of x , this means the x 2 = 3 or x 2 = − 2.

Quadratics with Substitution Sometimes we can transform complex equations into ones we know how to solve. Example. Solve the equation: x 4 − x 2 − 6 = 0 Solution. Substitute the variable u in place of x 2 . Note that since u = x 2 , it follows that u 2 = x 4 . ◮ The equation x 4 − x 2 − 6 = 0 becomes u 2 − u − 6 = 0. ◮ The equation u 2 − u − 6 = 0 can be solved by factoring as ( u − 3)( u + 2) = 0 to conclude that either u = 3 or u = − 2. ◮ In terms of x , this means the x 2 = 3 or x 2 = − 2. √ √ ◮ The former means that x = 3 or x = − 3, and the latter is √ √ impossible, so the solution set is { 3 , − 3 } .

Quadratics with Substitution Solve each of the following using a substitution. 1. (2 x + 3) 2 − 6(2 x + 3) + 8 = 0 2. x 6 − 72 x 3 + 512 = 0 3. 7 x − 2 − 6 x − 1 − 1 = 0 1 1 4. ( x + 6) 2 = x + 6 + 20

Linear Inequalities We solve linear inequalities using the same rules we used for linear equations. The only different “rule” is that multiplying/dividing both sides by a negative number reverses the inequality .

Linear Inequalities We solve linear inequalities using the same rules we used for linear equations. The only different “rule” is that multiplying/dividing both sides by a negative number reverses the inequality . Example . Solve − 2 x + 3 > x + 8

Linear Inequalities We solve linear inequalities using the same rules we used for linear equations. The only different “rule” is that multiplying/dividing both sides by a negative number reverses the inequality . Example . Solve − 2 x + 3 > x + 8 ◮ Subtract 3 to both sides: − 2 x > x + 5

Linear Inequalities We solve linear inequalities using the same rules we used for linear equations. The only different “rule” is that multiplying/dividing both sides by a negative number reverses the inequality . Example . Solve − 2 x + 3 > x + 8 ◮ Subtract 3 to both sides: − 2 x > x + 5 ◮ Add − x to both sides: − 3 x > 5

Linear Inequalities We solve linear inequalities using the same rules we used for linear equations. The only different “rule” is that multiplying/dividing both sides by a negative number reverses the inequality . Example . Solve − 2 x + 3 > x + 8 ◮ Subtract 3 to both sides: − 2 x > x + 5 ◮ Add − x to both sides: − 3 x > 5 ◮ Divide by − 3: x < − 5 / 3 (Note that the inequality reversed.)

Linear Inequalities We solve linear inequalities using the same rules we used for linear equations. The only different “rule” is that multiplying/dividing both sides by a negative number reverses the inequality . Example . Solve − 2 x + 3 > x + 8 ◮ Subtract 3 to both sides: − 2 x > x + 5 ◮ Add − x to both sides: − 3 x > 5 ◮ Divide by − 3: x < − 5 / 3 (Note that the inequality reversed.) ◮ As an interval, the solution set is ( −∞ , − 5 / 3).

Examples Solve each of the following inequalities, and express the solution set in interval notation. 1. 2 x + 1 < 5 2. − 3 x + 4 ≤ − x − 3 3. 1 4 x + 1 2 ≥ − 1 4

Compound Inequalities Compound linear inequalities follow the same rules, but the strategy is to isolate x in the middle inequality.

Compound Inequalities Compound linear inequalities follow the same rules, but the strategy is to isolate x in the middle inequality. Example. Solve 1 ≤ 4 x + 3 ≤ 5

Compound Inequalities Compound linear inequalities follow the same rules, but the strategy is to isolate x in the middle inequality. Example. Solve 1 ≤ 4 x + 3 ≤ 5 Solution. Subtract 3 from each term: − 2 ≤ 4 x ≤ 2