27

Section 1.5 - Other Types of Equations

The numbers of solutions to a polynomial with n degree, where n is Natural Number, are n solutions.

Solving a Polynomial Equation by factoring

Example

Solve: 4x4 = 12x

2

Solution

4x4 - 12x

2 = 0

4x2 (x

2 – 3) = 0

24 0x

2 3 0x

2 0x

2 3x

, 0 0x

3x

Example

Solve: 2 x3 + 3 x

2 = 8x + 12

Solution

3 22 3 12 08x x x

2 2 3 4 2 3 0x x x

22 3 4 0x x

2 3 0x

2 4 0x

2 3x

2 4x

32

x

24x

28

Solving a Radical Equation

Power Property

If P and Q are algebraic expressions, then every solution of the equation P Q is also a solution of the

equation n nP Q ; for any positive integer n.

Example

Solve 15 2 0x x

Solution

15 2x x

22 15 2x x

2 15 2x x

2 2 15 0x x

( 3)( 5) 0x x

3 0x 5 0x

3x 5x

Check

3x 5x

15 2 ) 03 (3 15 25 5 0( )

3 9 0 5 25 0

3 3 0 ( )true 5 5 0 ( )false

3x is the only solution

29

Solving Radical Equations of the Form

mn

x k

Assume that m and n are positive integers

If m is even:

mn

x k

nm

nm

mn

x k

nm

kx

If m is odd:

mn

x k

nm

nm

mn

x k

nm

kx

Example

Solve:

a)

32

5 25 0x

32

5 25x

32 25 5

5x

23

5x

3 25x

3 25

b)

23

8 4x

23

4x

3/2

4x

34

4 4

8

30

Equations that Are Quadratic in Form

2 0x xa b c

12

0x xa b c

12

0u ua b c

12

0n nx xa b c

2 0u ua b c

Example

Solve: 065 24 xx

Solution

065 222 xx 2 2 2 3 0x x or

0652 UU 2

2 0x

23 0x

0652 UU 2

2x

23x

Solve for U 2x

3x

)1(2

)6)(1(42)5()5( U

2

24255

2

15

5 12

5 12

2

3

U

U

2x U

2

2

2 2

3 3

x x

x x

31

Example

Solve: 2/3 1/31 1 2 0x x

Solution

1/31u x

1/3 1/3

1 2 1 1 0x x

2 2 0u u

( 2)( 1) 0u u

2 0u 1 0u

2u 1u

1/31 2u x 1/3

1 1u x

31 2x 31 1x

1 8x 1 1x

7x 2x

Example

Solve: 04113 3/12/3 xx Or factor

04113 3/123/1 xx 1/3 1/33 1 4 0x x

04113 2 UU 1/33 1 0x

1/3 4 0x

04113 2 UU

Solve for U

U = 2( 11) 11 4(3)( 4)

2(3)

61311

1/3x U 1/3 11 13

6x

1/3 11 136

x

13

4

3

13

x

34x

127

64

32

Solving an Absolute Value Equation

If c is a positive real number and X represents any algebraic expression, then |X| = c is equivalent to

X = c or X = - c

|X| = c X = c or X = c

Properties of Absolute Value

1. For b > 0, a b if and only if (iff) a b or a b

2. a b iff a b or a b

For any positive number b:

3. a b iff b a b

4. a b iff a b or a b

Example

Solve: |2x – 1| = 5

2x – 1 = 5 2x – 1 = -5

2x = 6 2x = -4

x = 3 x = -2 Solutions: x = -2, 3

Example

Solve: 4|1 - 2x| - 20 = 0

4|1 - 2x| = 20

|1 - 2x| = 5

1 - 2x = 5 1 - 2x = -5

-2x = 4 -2x = -6

x = -2 x = 3 Solutions: x = -2, 3

33

Exercise Section 1.5 - Other Types of Equations

1. Solve: 3 23 2 12 8x x x

2. Solve 4 23 10x x

3. Solve: 45 40x x

4. Solve 4 29 9 2 0x x

5. Solve: 4 2720 89x x

6. Solve 2 3 0x x

7. Solve: 3 3x x

8. Solve: 11 1x x

9. Solve: 2 3 2 1x x

10. Solve: 2 3 1 1x x

11. Solve: 5 3 2x x

12. Solve: 3 2 34 4 1 0x x x

13. Solve 4 212 11 2 0x x

14. Solve 4 22 7 5 0x x

15. Solve: 4 2

5 4 0x x

16. Solve 4 23 10x x

17. Solve 3 4 0x x

18. Solve 1/4

25 6x x

19. Solve 1/4

2 24 3x x

20. Solve: 5/2 32x

21. Solve

7 5 2 16x

22. Solve

34

4 1 7 10x

23. Solve 7 6 2x

24. Solve equation: 5 3 12x

34

25. Solve equation: 4 2 5x

26. Solve equation: 6 1 31

xx

27. Solve equation: 1 1 3x x

28. Two vertical poles of lengths 4 feet and 10 feet stand 15 feet apart. A cable reaches from the top of

one pole to some point on the ground between the poles and then to the top of the other pole. Where

should this point be located to use 24 feet of cable?

29. Towns A and B are located 6 miles and 3 miles, respectively, from a major expressway. The point

on the expressway closet to town A is 12 miles from the point on the expressway closet to town B.

Two new roads are to be built from A to the expressway and then to B.

a) Express the combined lengths of the new road in terms of x.

b) If the combined lengths of the new roads is 15 miles, what distance does x represent?

L2

L1

40

Solution Section 1.5 - Other Types of Equations

Exercise

Solve 3 23 2 12 8x x x

Solution

3 23 2 (12 8) 0x x x

2 3 2 4(3 2) 0x x x

23 2 4 0x x

3 2 0x 2 4 0x

3 2x 2 4x

23

x 4 2x

Exercise

Solve 4 23 10x x

Solution

4 23 10 0x x

2 25 2 0x x

2 5 0x 2 2 0x

2 5x 2 2x

5x 2x

5x i

Exercise

Solve: 45 40x x

Solution

45 40 0x x

35 8 0x x

3 35 2 0x x

41

2 25 2 2 2 0x x x x

2 25 02 4xxx x

20 00 2 42 x xxx

22 2 4(1)(

24)

2(1)xx

2 12

2

2 2 3

2

i

2 1 3

2

i

1 3x i

Exercise

Solve 4 29 9 2 0x x

Solution

Assume 2 4 2x u x u

Then we can rewrite the equation in a quadratic form: 29 9 2 0u u

Solve for u using the quadratic formula.

29 9 4 9 2

2 9u

2 4

2

b b ac

a

9 9

18

9 3

18

9 3 6 118 18 3

9 3 12 218 18 3

u

u

Since 2u x

2

2

13

2

1

3

32

3

x u

x u

x

x

1

3

2

3

3

3

3

33

633

x

x

42

Exercise

Solve: 4 2720 89x x

Solution

4 289 720 0x x

2 1

2 289 720 0x x

Assume: 2u x

2 89 720 0u u

289 89 4 1 720

2 1u

89 5041

2

89 71

2

2

2

89 719

2

89 7180

2u

x

x

u

2 39 xx

2 80 80 16 4 55x x x

Exercise

Solve 2 3 0x x

Solution

2 3x x

22

2 3x x

2 2 3x x 2 2 3 0x x

( 1)( 3) 0x x

1 0x 3 0x

1x 3x

Check

1x 3x

2 31) 0( ( 1) (3) (32 3) 0

43

1 1 0 3 9 0

False True

3x is the only solution

Exercise

Solve: 3 3x x

Solution

33 xx

2 2 33 xx

23 6 9x x x 2 7 6 0x x

2( 7) ( 7) 4(1)

(1)

(6)

2x

7 49 24

2

7 25

2

7 5

2

7 5 12

2 2

7 5 2

2 2

6

1

x = 1, 6

Check:

x = 1 1331 5 = 1 (Not a solution)

x = 6 6336 6 = 6 x = 6 is the only solution

Exercise

Solve 11 1x x

Solution

xx 111 Square both side

22

111 xx 2 222a b a ab b

2111 2 xx x

20 2 1 11x xx

44

20 3 10x x 2 3 10 0x x Solve for x

x = 5, 2

Check

x = 5 5 5 11 1 5 16 1 5 4 1 1 1

x = -2 2 2 11 1 2 9 1 2 3 1 5 1 ( )False

Solution: x = 5

Exercise

Solve: 2 3 2 1x x

Solution

2 3 1 2x x

2 2

2 3 1 2x x

2

1 2 2 22 3 x xx

2 22 3 1 2xx x

2 22 4 2 xx x

2 22 xx

22

2 2( 2) xx

24( )4 4 2x x x

244 4 8x x x

244 4 8 0x x x

22( 8) ( 8) 4(1)(12)4

2 2(1)

2

8 64 48 8 16 8 4

2 2 2

8 12 0

b b ac

a

x x

x

x = 2, 6

Check

2 2(2) 3 2 2 1 1 0 1x

6 2(6) 3 6 2 1 3 2 1 5 1x

Solution: x = 2

45

Exercise

Solve: 5 3 2x x

Solution

325 xx

2 32

2 5 xx

2 33445 xxx

33445 xxx

34345 xxx

344 x

31 x

1 = x – 3

x = 4

Check: 23454

3 – 1 = 2 (True statement)

x = 4 is a solution

46

Exercise

Solve: 2 3 1 1x x

Solution

2 2

2 3 1 1x x 2 222a b a ab b

2 3 1 2 1 1x x x

2 3 2 1 2x x x

1 2 1x x

22

1 2 1x x

2 2 1 4( 1)x x x

2 2 1 4 4x x x 2 2 3 0x x

( 3)( 1) 0x x

3 0x 1 0x

3x 1x

Check

3x 1x

2 3 1(3 3) 1) ( 2 3 1) ( ) 1( 1 1

9 1 4 1 1 0

3 3 ( )true 11 ( )true

3 1x and are solutions

Exercise

Solve: 5 3 2x x

Solution

325 xx

2 32

2 5 xx

2 33445 xxx

33445 xxx

34345 xxx

344 x

47

31 x

1 = x – 3

x = 4

Check: 23454

3 – 1 = 2 (True statement)

x = 4 is a solution

Exercise

Solve: 3 2 34 4 1 0x x x

Solution

3

3 2 334 4 1x x x

24 4 1x x x

24 5 1 0x x

(4 1)( 1) 0x x

4 1 0x 1 0x

14

x 1x

Check

14

x 1x

1 1

43

4

23 4

44 11 0 2 33 14 ( ) 14 11 0

1 13 34 4

0 3 3 14 4 1 0

0 0 ( )true 0 0 ( )true

Solutions: 14

1, x

48

Exercise

Solve 4 212 11 2 0x x

Solution

2

2 212 11 2 0x x

2 23 2 4 1 0x x

2 let u x

212 11 2 0u u

2( 11) ( 11) 4(12)(2)

2(12)u

(3 2)(4 1) 0u u 11 25

24

3 2 0u 4 1 0u 11 5

24

22

3u x

214

u x 2 11 5

24x

2 11 5

24x

2

3x

14

x 16

24

23

8

24

14

3

3

2

3x

12

x

6

3x

The solution set is 6 1, 3 2

Exercise

Solve 4 22 7 5 0x x

Solution

2 22 5 1 0x x

22 5 0x

2 1 0x

2 5

2x

2 1x

5

2x 1x

2

2

2

5x

10

2x

49

Exercise

Solve 4 2

5 4 0x x

Solution

045 222 xx 2 21 4 0x x

2

5 4 0U U 2

1 0x 2

4 0x

Solve for U 2

( 5) 5 4(1)(4)

2(1)U

21x

24x

5 92

1x

2x

5 32

2 5 3

22

2 5 3

2

1 1

4 2

x x

x U

x x

Exercise

Solve 4 23 10x x

Solution

4 23 10 0x x

2 25 2 0x x

2 5 0x 2 2 0x

2 5x 2 2x

5x 2x

5x i

Exercise

Solve 3 4 0x x

Solution

4 1 0x x

4 0x 1 0x

4x 1x Impossible

16x

50

Exercise

Solve 1/4

25 6x x

Solution

1/4

24

45 6x x

2 45 6x x

24 5 6 0x x

22 3 2 0x x

2 3 0x 2 2 0x

2 3x 2 2x

3x 2x

Exercise

Solve 1/4

2 24 3x x

Solution

4

44

1/2 24 3x x

2 24 81x x

2 24 81 0x x

( 27)( 3) 0x x

27 0x

3 0x

27x

3x

51

Exercise

Solve: 5/2 32x

Solution

2/532x Reciprocal or use the calculator: 32^(2/5) = 4

5 232x

51024

4

Exercise

Solve

7 5 2 16x

Solution

7 5 16 2x

7 5 14x

1457

x

5 2x

5 2x

25

x

5 2x

25

x

Solutions: 25

x

Exercise

Solve 7 6 2x

Solution

7 2 6x

47 x

No solution or , since the absolute value can't be a negative

52

Exercise

Solve

34

4 1 7 10x

Solution

3

44 1 10 7x

4 3 10 7x

Distribute 4

3

44 1 3x

4 3 3x

3

4

314

x

3 34 4

1 x

3 34 4

1x

3 14 4

x

1 44 3

x

13

x

3 34 4

1 x

3 34 4

1x

3 74 4

x

7 44 3

x

73

x

4 3 3x

73x

73

x

4 3 3x

13x

13

x

Solutions: 71 ,3 3

x

Exercise

Solve equation: 5 3 12x

Solution

5 3 12x 5 3 12x

53 12 55 x 5 3 125 5x

3 7x 3 17x

73

x 173

x

Exercise

Solve equation: 4 2 5x

Solution

4 2 5x

4 2 5x

4 7x

4 3x

74

x

34

x

53

Exercise

Solve equation: 6 1 31

xx

Solution

6 1 31

xx

6 1 31

xx

6 1( 1) (31

1)xx

x x

6 1 3( 1)x x

6 1 3 3x x

6 1 3 3x x

3 36 1 33xx xx

36 31 3 3xx xx

9 1 3x

3 1 3x

9 2x

3 4x

29

x

43

x

Exercise

Solve equation: 1 1 3x x

Solution

1 (1 3 )x x

1 1 3x x

1 1 3x x

113x x 3 1 1x x

4 0x

2 2x

0x

1x

Solution: 0, 1

54

Exercise

Two vertical poles of lengths 4 feet and 10 feet stand 15 feet apart. A cable reaches from the top of one

pole to some point on the ground between the poles and then to the top of the other pole. Where should

this point be located to use 24 feet of cable?

Solution

2 2 21

4l x 1

2 16l x

22 22

15 10l x 2

215 100l x

1 224l l

22 16 15 100 24x x

2 215 100 24 16x x Square both sides

2 2

2 215 100 24 16x x

2 2 230 225 100 576 48 16 16x x x x

22 257630 325 48 116 6x x xx

230 267 48 16x x

230 267 48 16x x

2 2 230 267 48 16x x

2 2900 16020 71289 2304 16x x x

2 2900 16020 71289 2304 36864x x x

2 2900 16020 7128 239 004 36864xx x

21404 16020 34425 0x x

Solve for x:

x = 13.259

L2

L1

55

Exercise

Towns A and B are located 6 miles and 3 miles, respectively, from a major expressway. The point on the

expressway closet to town A is 12 miles from the point on the expressway closet to town B. Two new

roads are to be built from A to the expressway and then to B.

a. Express the combined lengths of the new road in terms of x.

b. If the combined lengths of the new roads is 15 miles, what distance does x represent?

Solution

1d 2

d

a) 1

2 2 2 21

6 36d x d x

2

2 22 22

12 3 12 9d x d x

1 2

22 36 12 9d d x x

b) 22 36 12 9 15x x

2 236 15 144 24 9x x x

22

2236 15 24 153x x x

2 2 236 225 30 24 153 24 153x x x x x

230 24 153 24 342x x x

2

2230 24 153 24 342x x x

2 2900 24 153 576 16416 116964x x x x

2 2900 21600 137700 576 16416 116964x x x x

2324 5184 20736 0x x

Solve for x: 8x