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

SLIDE 1

College Algebra - Sections 1.4-1.6

Dr. Doug Ensley

Shippensburg University April 20, 2015

SLIDE 2

Radicals

The basic strategy for solving equations including radicals (such as √ ,

3

√ ,

4

√ , etc.:

1. Isolate the radical part of the equation. If there is more than
ne, 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.

SLIDE 3

Radicals

The basic strategy for solving equations including radicals (such as √ ,

3

√ ,

4

√ , etc.:

1. Isolate the radical part of the equation. If there is more than
ne, 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√2x + 1 − 7 = 8

SLIDE 4

Radicals

The basic strategy for solving equations including radicals (such as √ ,

3

√ ,

4

√ , etc.:

1. Isolate the radical part of the equation. If there is more than
ne, 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√2x + 1 − 7 = 8

◮ Solution. Add 7 to both sides: 3√2x + 1 = 15

SLIDE 5

Radicals

The basic strategy for solving equations including radicals (such as √ ,

3

√ ,

4

√ , etc.:

1. Isolate the radical part of the equation. If there is more than
ne, 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√2x + 1 − 7 = 8

◮ Solution. Add 7 to both sides: 3√2x + 1 = 15 ◮ Divide both sides by 3: √2x + 1 = 5. (Isolates the radical.)

SLIDE 6

Radicals

The basic strategy for solving equations including radicals (such as √ ,

3

√ ,

4

√ , etc.:

1. Isolate the radical part of the equation. If there is more than
ne, 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√2x + 1 − 7 = 8

◮ Solution. Add 7 to both sides: 3√2x + 1 = 15 ◮ Divide both sides by 3: √2x + 1 = 5. (Isolates the radical.) ◮ Square both sides: 2x + 1 = 25.

SLIDE 7

Radicals

The basic strategy for solving equations including radicals (such as √ ,

3

√ ,

4

√ , etc.:

1. Isolate the radical part of the equation. If there is more than
ne, 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√2x + 1 − 7 = 8

◮ Solution. Add 7 to both sides: 3√2x + 1 = 15 ◮ Divide both sides by 3: √2x + 1 = 5. (Isolates the radical.) ◮ Square both sides: 2x + 1 = 25. ◮ Continue until you’ve solved for x. Answer: x = 12

SLIDE 8

Radicals

The basic strategy for solving equations including radicals (such as √ ,

3

√ ,

4

√ , etc.:

1. Isolate the radical part of the equation. If there is more than
ne, 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√2x + 1 − 7 = 8

◮ Solution. Add 7 to both sides: 3√2x + 1 = 15 ◮ Divide both sides by 3: √2x + 1 = 5. (Isolates the radical.) ◮ Square both sides: 2x + 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

SLIDE 9

Examples

Solve each of the following for x:

1. √2x − 7 = 5
2. 3 +

√ x2 + 5x = 9

3. 4 + √5x − 6 = x

4. √ x2 − x − 6 = x + 3

SLIDE 10

More Examples

Solve each of the following for x: 1.

5

√ x2 + 4x = 2

2. (5x + 6)1/2 = 9
3. (3x + 3)1/3 = −3
4. √3x − 5 − √x + 7 = 2

SLIDE 11

Quadratics with Substitution

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

SLIDE 12

Quadratics with Substitution

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

Example. Solve the equation:

x4 − x2 − 6 = 0

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Quadratics with Substitution

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

Example. Solve the equation:

x4 − x2 − 6 = 0

Solution. Substitute the variable u in place of x2. Note that since

u = x2, it follows that u2 = x4.

SLIDE 14

Quadratics with Substitution

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

Example. Solve the equation:

x4 − x2 − 6 = 0

Solution. Substitute the variable u in place of x2. Note that since

u = x2, it follows that u2 = x4.

◮ The equation x4 − x2 − 6 = 0 becomes u2 − u − 6 = 0.

SLIDE 15

Quadratics with Substitution

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

Example. Solve the equation:

x4 − x2 − 6 = 0

Solution. Substitute the variable u in place of x2. Note that since

u = x2, it follows that u2 = x4.

◮ The equation x4 − x2 − 6 = 0 becomes u2 − u − 6 = 0. ◮ The equation u2 − u − 6 = 0 can be solved by factoring as

(u − 3)(u + 2) = 0 to conclude that either u = 3 or u = −2.

SLIDE 16

Quadratics with Substitution

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

Example. Solve the equation:

x4 − x2 − 6 = 0

Solution. Substitute the variable u in place of x2. Note that since

u = x2, it follows that u2 = x4.

◮ The equation x4 − x2 − 6 = 0 becomes u2 − u − 6 = 0. ◮ The equation u2 − 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 x2 = 3 or x2 = −2.

SLIDE 17

Quadratics with Substitution

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

Example. Solve the equation:

x4 − x2 − 6 = 0

Solution. Substitute the variable u in place of x2. Note that since

u = x2, it follows that u2 = x4.

◮ The equation x4 − x2 − 6 = 0 becomes u2 − u − 6 = 0. ◮ The equation u2 − 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 x2 = 3 or x2 = −2. ◮ The former means that x =

√ 3 or x = − √ 3, and the latter is impossible, so the solution set is { √ 3, − √ 3}.

SLIDE 18

Quadratics with Substitution

Solve each of the following using a substitution.

1. (2x + 3)2 − 6(2x + 3) + 8 = 0
2. x6 − 72x3 + 512 = 0
3. 7x−2 − 6x−1 − 1 = 0

4. 1 (x + 6)2 = 1 x + 6 + 20

SLIDE 19

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.

SLIDE 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.

Example. Solve −2x + 3 > x + 8

SLIDE 21

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 −2x + 3 > x + 8

◮ Subtract 3 to both sides: −2x > x + 5

SLIDE 22

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 −2x + 3 > x + 8

◮ Subtract 3 to both sides: −2x > x + 5 ◮ Add −x to both sides: −3x > 5

SLIDE 23

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 −2x + 3 > x + 8

◮ Subtract 3 to both sides: −2x > x + 5 ◮ Add −x to both sides: −3x > 5 ◮ Divide by −3: x < −5/3 (Note that the inequality reversed.)

SLIDE 24

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 −2x + 3 > x + 8

◮ Subtract 3 to both sides: −2x > x + 5 ◮ Add −x to both sides: −3x > 5 ◮ Divide by −3: x < −5/3 (Note that the inequality reversed.) ◮ As an interval, the solution set is (−∞, −5/3).

SLIDE 25

Examples

Solve each of the following inequalities, and express the solution set in interval notation.

1. 2x + 1 < 5
2. −3x + 4 ≤ −x − 3
3. 1

4x + 1 2 ≥ −1 4

SLIDE 26

Compound Inequalities

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

SLIDE 27

Compound Inequalities

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

Example. Solve 1 ≤ 4x + 3 ≤ 5

SLIDE 28

Compound Inequalities

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

Example. Solve 1 ≤ 4x + 3 ≤ 5
Solution. Subtract 3 from each term: −2 ≤ 4x ≤ 2

SLIDE 29

Compound Inequalities

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

Example. Solve 1 ≤ 4x + 3 ≤ 5
Solution. Subtract 3 from each term: −2 ≤ 4x ≤ 2

Divide each each term by 4: − 1

2 ≤ x ≤ 1 2

SLIDE 30

Compound Inequalities

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

Example. Solve 1 ≤ 4x + 3 ≤ 5
Solution. Subtract 3 from each term: −2 ≤ 4x ≤ 2

Divide each each term by 4: − 1

2 ≤ x ≤ 1 2

So the solution is the interval

− 1

2, 1 2

SLIDE 31

Compound Inequalities

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

Example. Solve 1 ≤ 4x + 3 ≤ 5
Solution. Subtract 3 from each term: −2 ≤ 4x ≤ 2

Divide each each term by 4: − 1

2 ≤ x ≤ 1 2

So the solution is the interval

− 1

2, 1 2

Don’t forget that if you multiply/divide by a negative number, you

must reverse the inequalities.

SLIDE 32

Examples

Solve each of the following; express the answer as an interval.

1. −1 ≤ 3x − 1 ≤ 8
2. 22 > 9 − 3x > 3
3. −2 < 2x − 4

5 < 0

SLIDE 33

Other Inequalities

Example. Solve 2x2 + 3 > 3x + 8

SLIDE 34

Other Inequalities

Example. Solve 2x2 + 3 > 3x + 8
1. Change the problem to be an equation: 2x2 + 3 = 3x + 8

SLIDE 35

Other Inequalities

Example. Solve 2x2 + 3 > 3x + 8
1. Change the problem to be an equation: 2x2 + 3 = 3x + 8
2. Rewrite to get 0 alone: 2x2 − 3x − 5 = 0

SLIDE 36

Other Inequalities

Example. Solve 2x2 + 3 > 3x + 8
1. Change the problem to be an equation: 2x2 + 3 = 3x + 8
2. Rewrite to get 0 alone: 2x2 − 3x − 5 = 0
3. Solve by factoring: x = −1 or x = 5/2

SLIDE 37

Other Inequalities

Example. Solve 2x2 + 3 > 3x + 8
1. Change the problem to be an equation: 2x2 + 3 = 3x + 8
2. Rewrite to get 0 alone: 2x2 − 3x − 5 = 0
3. Solve by factoring: x = −1 or x = 5/2
4. Plot these open points on a number line:

SLIDE 38

Other Inequalities

Example. Solve 2x2 + 3 > 3x + 8
1. Change the problem to be an equation: 2x2 + 3 = 3x + 8
2. Rewrite to get 0 alone: 2x2 − 3x − 5 = 0
3. Solve by factoring: x = −1 or x = 5/2
4. Plot these open points on a number line:
5. Sample around these points to create a sign chart:

SLIDE 39

Other Inequalities

Example. Solve 2x2 + 3 > 3x + 8
1. Change the problem to be an equation: 2x2 + 3 = 3x + 8
2. Rewrite to get 0 alone: 2x2 − 3x − 5 = 0
3. Solve by factoring: x = −1 or x = 5/2
4. Plot these open points on a number line:
5. Sample around these points to create a sign chart:
6. Final answer: (−∞, −1) ∪ (2.5, ∞)

SLIDE 40

Absolute Value – Equations

Example. Solve 3 · |4x + 5| − 2 = 7

SLIDE 41

Absolute Value – Equations

Example. Solve 3 · |4x + 5| − 2 = 7
1. Isolate the absolute value part. Don’t make shit up.

SLIDE 42

Absolute Value – Equations

Example. Solve 3 · |4x + 5| − 2 = 7
1. Isolate the absolute value part. Don’t make shit up.
2. Add 2 to both side & divide by 3: |4x + 5| = 3

SLIDE 43

Absolute Value – Equations

Example. Solve 3 · |4x + 5| − 2 = 7
1. Isolate the absolute value part. Don’t make shit up.
2. Add 2 to both side & divide by 3: |4x + 5| = 3
3. Remove absolute value in favor of creating TWO problems:

4x + 5 = 3 OR 4x + 5 = −3

SLIDE 44

Absolute Value – Equations

Example. Solve 3 · |4x + 5| − 2 = 7
1. Isolate the absolute value part. Don’t make shit up.
2. Add 2 to both side & divide by 3: |4x + 5| = 3
3. Remove absolute value in favor of creating TWO problems:

4x + 5 = 3 OR 4x + 5 = −3

4. Solve both: x = −1/2 OR x = −2

SLIDE 45

Absolute Value – Equations

Example. Solve 3 · |4x + 5| − 2 = 7
1. Isolate the absolute value part. Don’t make shit up.
2. Add 2 to both side & divide by 3: |4x + 5| = 3
3. Remove absolute value in favor of creating TWO problems:

4x + 5 = 3 OR 4x + 5 = −3

4. Solve both: x = −1/2 OR x = −2
5. CHECK: 3 · |4(−1/2) + 5| − 2 = 3 · |3| − 2 = 7.

CHECK: 3 · |4(−2) + 5| − 2 = 3 · | − 3| − 2 = 7.

SLIDE 46

Absolute Value – Equations

Example. Solve 3 · |4x + 5| − 2 = 7
1. Isolate the absolute value part. Don’t make shit up.
2. Add 2 to both side & divide by 3: |4x + 5| = 3
3. Remove absolute value in favor of creating TWO problems:

4x + 5 = 3 OR 4x + 5 = −3

4. Solve both: x = −1/2 OR x = −2
5. CHECK: 3 · |4(−1/2) + 5| − 2 = 3 · |3| − 2 = 7.

CHECK: 3 · |4(−2) + 5| − 2 = 3 · | − 3| − 2 = 7.

6. So the solution set is {−2, −1/2}.

SLIDE 47

Examples

Solve each of the following:

1. |2x + 1| = 5
2. |1 − 2t| + 5 = 14

3.

x

5 + 1 2

= 2
4. |x2 − 3x| = 18

SLIDE 48

Absolute Value – Inequalities

Example. Solve |4x + 1| − 7 ≤ 4

SLIDE 49

Absolute Value – Inequalities

Example. Solve |4x + 1| − 7 ≤ 4
1. Change the problem to be an equation: |4x + 1| − 7 = 4

SLIDE 50

Absolute Value – Inequalities

Example. Solve |4x + 1| − 7 ≤ 4
1. Change the problem to be an equation: |4x + 1| − 7 = 4
2. Isolate absolute value: |4x + 1| = 11

SLIDE 51

Absolute Value – Inequalities

Example. Solve |4x + 1| − 7 ≤ 4
1. Change the problem to be an equation: |4x + 1| − 7 = 4
2. Isolate absolute value: |4x + 1| = 11
3. Replace with two equations: 4x + 1 = 11 OR 4x + 1 = −11

SLIDE 52

Absolute Value – Inequalities

Example. Solve |4x + 1| − 7 ≤ 4
1. Change the problem to be an equation: |4x + 1| − 7 = 4
2. Isolate absolute value: |4x + 1| = 11
3. Replace with two equations: 4x + 1 = 11 OR 4x + 1 = −11
4. Solve (x = 5/2 OR x = −3) and plot these closed points on

a number line:

SLIDE 53

Absolute Value – Inequalities

Example. Solve |4x + 1| − 7 ≤ 4
1. Change the problem to be an equation: |4x + 1| − 7 = 4
2. Isolate absolute value: |4x + 1| = 11
3. Replace with two equations: 4x + 1 = 11 OR 4x + 1 = −11
4. Solve (x = 5/2 OR x = −3) and plot these closed points on

a number line:

5. Sample around these points to create a sign chart:

SLIDE 54

Absolute Value – Inequalities

Example. Solve |4x + 1| − 7 ≤ 4
1. Change the problem to be an equation: |4x + 1| − 7 = 4
2. Isolate absolute value: |4x + 1| = 11
3. Replace with two equations: 4x + 1 = 11 OR 4x + 1 = −11
4. Solve (x = 5/2 OR x = −3) and plot these closed points on

a number line:

5. Sample around these points to create a sign chart:
6. Final answer: [−3, 2.5]

SLIDE 55

Examples

Solve each of the following:

1. |2x + 1| < 5
2. |1 − 2t| + 5 ≥ 14

3.

x

5 + 1 2

> 2
4. |x2 − 3x| ≤ 18