Unit 10 ~ Contents

Unit 10 ~ Contents 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11 10.12 10.13 10.14 10.15 10.16

Algebra Beauty and Awe ~ Maury . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Quadratic Equations:The Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Fractional Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Absolute Value Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Canceling in Unit Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Quiz 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Paths in the Seas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Permutation Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Graphing Absolute Value Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 The Discriminant of a Quadratic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Multiplying and Dividing With Fractional Exponents. . . . . . . . . . . . . . . . . . . . . . . . . 26 Quiz 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Just One More Foot, Please . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Longer Unit Conversions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Applications Using Inequalities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Function Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 A System With an Equation and an Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Review for Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Four-Color Maps—Cartography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

10.1

Quadratic Equations: The Quadratic Formula

The standard form of a quadratic equation can be solved to find the value of x. When this is done using the method of completing the square, the standard form of quadratic equations is transformed into the quadratic formula. Derivation of the Quadratic Formula ax 2 + bx + c = 0

ax 2 + bx + c a x2 +

x2 +

=

Quadratic equation in standard form.

0 a

Both sides divided by a, the coefficient of x 2.

b c x+ = 0 a a

Simplified.

x2 +

Constant term moved to the right side.

b c x = – a a

b2 b x + 4a2 = a (x + (x +

b 2 ) = 2a

b ) = 2a

x+

b = 2a

b2 c 4a2 – a

2 ~ Algebra I Unit 10

b2 4a2 added (to both sides) to complete the square.

b2 – 4ac 4a2

± ±

x = – x =

To find the term to complete the square, divide the coefficient of the middle term by 2 and square it. b2 b b b ÷2= and ( )2 = 4a2 a 2a 2a

√

Trinomial on left factored, right side simplified.

b2 – 4ac 4a2

√b2 – 4ac

b 2a

2a

±

√b2 – 4ac 2a

–b ± √b2 – 4ac 2a

Square root of both sides taken. Right side simplified. b moved to right side. 2a Simplified (two fractions combined).

The Quadratic Formula. Memorize it! x =

–b ±√b2 – 4ac 2a

for equations in the form ax 2 + bx + c = 0 (x ≠ 0)

The quadratic formula can be used to solve a quadratic equation without using all the steps required by other methods. To solve quadratic equations using the quadratic formula, substitute the values of a, b, and c from the original equation into the quadratic formula and simplify. Example 1. Use the quadratic formula to solve the equation 6x 2 – 11x – 2 = 0. x = x = x = x =

– b ± √b2 – 4ac 2a

(In this equation a = 6, b = –11, and c = – 2.)

–(–11) ± √(–11)2 – 4(6)(– 2) 2(6) 11 ± √169 12 11 ± 13 12

x = 2, –

1 6

The quadratic formula. The a, b, and c values substituted. Simplified. Simplified. Solutions.

Note: The equation in this example could be solved by factoring, but since the coefficient of the first term is not a 1, it is more easily done with the quadratic equation. Example 2. Use the quadratic formula to solve the equation 5x 2 – 7x + 1 = 0. x = x = x =

– b ± √b2 – 4ac 2a

–(–7) ± √(–7)2 – 4(5)(1) 2(5) 7 ± √29 10

(In this equation a = 5, b = –7, and c = 1.) The quadratic formula.

The a, b, and c values substituted. Solution.

10.1 Quadratic Equations: The Quadratic Formula ~ 3

Solve by using the quadratic formula. 10.1 1. 3x 2 + 2x – 1 = 0

2. x 2 + 7x – 18 = 0

3. 4x 2 – 15x + 9 = 0

4. 10x 2 – 13x – 3 = 0

5. 4x 2 + 7x – 2 = 0

6. 2x 2 + 9x + 2 = 0

Review

Solve. 9.14 7.

x–2 x–3 = x+1 x–4

8.

Simplify the expressions. 6.11 10.

2–x x2 – 4

11.

2x x 5 – = 3 4 3 x 2 – 3x x + 2x – 15 2

Factor polynomials completely. 8.2 13. 2x 3y + 5x 2y – 3xy

14. 3xy + 9x – 5y – 15

Divide the rational expressions. 7.7 x+2 5x + 10 16. 3x 2 ÷ 6x

17.

x+5 x 2 – 7x + 12 2 ÷ 2 x + 3x – 10 x – 5x + 6

9.

12.

2x + 1 x – 2 = –3 –9 12xy 2 42y 5z

15. 3r 2s4 – 48r 2

18.

z 5x 2z 3 ÷ 4xy 2 2y

Solve by using direct variation. 7.14

19. If y varies directly as the square of x, and y is 14 when x is 2, what is y when x is 7? 20. If b varies directly as a, and b is 9.1 when a is 0.7, what is b when a is 7?

21. Distance varies directly as the time. If the distance is 5.25 miles when time is 1.5 hours, what is the distance when the time is 0.5 hours?

Simplify the expressions. 9.11 22.

4x 2 4 2x 5 3y 2

23.

y–3 y+2

5y – 15 4

Solve the quadratic equations by completing the square. 9.12 25. x 2 – 2x – 1 = 0


26. x 2 – 8x + 5 = 0

24.

x–

1 2

1 x + 3 4

27. 3x 2 – 6x – 9 = 0

Solve for the requested information. 7.12

28. At the beginning of their 10th grade school year, Eric and Vance each decided to open a savings account (2.5% rate) dedicated to attending Bible School. After three years, at the beginning of Bible School, Vance’s average principal was $150 more than Eric’s and their combined interest was $101.25. What were their average principals?

Solve the systems of equations, using the method of your choice. 6.4, 6.12, 7.2 29. x – 2y = 9

x – 10y = 25

30. 2x – 5y = – 31

5x + 2y = – 34

Solve by using the quadratic formula. 10.1 32. 5x 2 + 7x + 2 = 0

33. 7x 2 + 8x + 1 = 0

31. 2x – 3y = 3

2x + 5y = – 21

34. 5x 2 + 3x – 9 = 0

10.1 Quadratic Equations: The Quadratic Formula ~ 5

10.2

Fractional Exponents

Normally roots are expressed using radical signs. However, fractional notation can also be used. The two terms below, one expressed using the radical sign and the other using a fractional exponent, are equal. a

√x a = x b

b

When a radical is expressed using a fraction exponent, the power of the radicand is the numerator and the index in the radical is the denominator. Exponent is numerator

Index is denominator

2

3

2

3

√x 2 = x ?

√x 2 = x 3

When there is no index, it is understood to be 2.

Sometimes a fractional exponent can be 2

√x2 = x4

4

=

1 2

1

4

√ 84 = 8 2

= x2

=2

= 82 = 64

Note how to convert to fractional notation when there are two or more variables. 1

1

1

√ xy = (xy) 2 or x 2 y 2

3

5

2

√x 3y 2 = x 5 y 5

Change from radical to fractional notation. Simplify if possible. 10.2 1. √7

5. 3√26

2. 3√100 6. √b5

3. √x 2 3 4

7. √y 3

4. √x 3 5

8. √162

The concept of fractional exponents works both ways. Radicals can be expressed using fractional exponents, and terms with fractional exponents can be expressed with radical signs. 2

3

x 3 = √x 2

1

3

8 3 = √8 = 2

Change from fractional notation to radical notation. Simplify if possible. 3

9. x 5

2

13. 8 3


1

10. 7 2 7

14. x 2

1

11. e 5 3

15. 3 4

2

12. y 3 4

16. y 3

Review Simplifying Fractions When Using the Quadratic Formula After solving an equation using the quadratic formula, and simplifying the radical, see if the solution is in lowest terms. Example 1. Use the quadratic formula to solve the equation 5x 2 – 6x – 1 = 0. x = x = x = x = x =

– b ± √b2 – 4ac 2a

(In this equation a = 5, b = – 6, and c = –1.) The quadratic formula.

–(– 6) ± √(– 6)2 – 4(5)(–1) 2(5) 6 ± √56 10

=

The a, b, and c values substituted.

6 ± 2√14 10

Radical simplified.

6 ± 2√14 3 ± √14 = 10 5

Fraction simplified. Numerator and denominator divided by 2.

3 ± √14 5

Solution.

Solve by using the quadratic formula. 10.1 17. x 2 + 13x + 22 = 0

Solve. 9.14 20.

24 3 1 – = 2 2 x

18. x 2 – 8x + 9 = 0

21.

3 3 = 4x – 15 20 – x

19. 9x 2 + 6x – 4 = 0

22. x – 1 +

4x – 1 x–1 = 4 2

Use a system of equations to solve the problems. 8.13

23. What are the values of two numbers whose average is 90 and whose difference is 24?

24. During one leap year Katrina’s weather records showed there were 68 more sunny days than cloudy. How many of each were there?

Divide. 9.9

25. (x 3 + 8) ÷ (x + 2)

26. (15x 3 – 19x 2 + 4) ÷ (5x + 2)

10.2 Fractional Exponents ~ 7

Combine like radicals. 7.9

3

27. 3 √18 + 5 √2

3

28. √16 – √2

29. √3 + √7

Solve by using inverse variation. 8.3

30. If y varies inversely as x 2, and y is 0.375 when x is 2, what is y when x is 3?

31. The speed of a gear (in rpm) varies inversely as the number of teeth in the gear. If a gear with 40 teeth rotates at a speed of 125 rpm, how fast would a gear with 50 teeth rotate? 32. If q varies inversely as p, and q is 21 when p is 0.5, what is q when p is 0.2?

Add or subtract. Leave answers in factored form. 9.4 33.

3y 5 + 4x 2y

34.

3h 4 – h2 – 4 h–2

Graph these systems of inequalities. 8.7 1 1 36. y ≥ – x – 3 and y ≤ x – 1 8 4 Divide. 9.1 38.

(6x 3 – 13x 2 + 19x – 5) (3x 2 – 5x + 7)

39.

37. y >

35.

2 5 x + 1 and x ≤ – 3 2

(12x 3 + 15x 2 + 2x – 15) (4x – 3)

40.

Change from radical to exponential notation. Simplify if possible. 10.2 41. √8

3

42. √27

43. √y

2


3

46. x 2

2

47. y 5

(24x 4 + 10x 3 – 31x 2 + 23x – 15) (6x – 5)

6

44. √n4

Change from exponential to radical notation. Simplify if possible. 10.2 45. x 7

m–1 m+1 – m–2 m+3

2

48. 8 3

10.3

Absolute Value Equations

The solution for x in the equation |x|= 9 is 9 and – 9. In the equation, |2b + 4| = 16, the quantity inside the absolute value bars also has a positive and a negative value. The positive value and the negative value can be found by setting up two equations. One equation with the quantity inside the absolute value bars equal to 16 and the other with the quantity equal to –16.

Solutions to an absolute value equation such as |x|= 9 are often written x = ±9.

Example 1. Solve the equation | 2b + 4 | = 16. | 2b + 4 | = 16

2b + 4 = 16

2b + 4 = –16

2b = 12

2b = – 20

Check:

b=6

| 2b + 4 | = 16

b = –10

| 2b + 4 | = 16

| 2(6) + 4 | = 16

| 2(–10) + 4 | = 16

16 = 16

16 = 16

| 16 | = 16

| –16 | = 16

Original equation. The expression 2b + 4 can equal +16 or –16 to make the equation true. Equation rewritten for both possible values of the expression. 4 subtracted from both sides.

Both sides divided by 2. Original equation.

Two different values for b substituted. Simplified.

Both equations true.

Both 6 and –10 cause the expression 2b + 4 to have an absolute value of 16 and, thus, satisfy the equation.

10.3 Absolute Value Equations ~ 9

Because absolute value equations require two values, the steps to solving equations must be modified to fit their situation. Steps to Solve Absolute Value Equations

1. Simplify any expressions inside and outside of the absolute value sign. 2. Isolate the absolute value expression on one side of the equation.

3. Use the expression inside the absolute value symbol to form two equations, one equal to the positive value and one equal to the negative value of the equation. 4. Solve both equations.

5. Check the answers by substituting them in the original equation. Example 2. Solve the equation | 4a – 3 – 2a| – 5 = 18. | 4a – 3 – 2a| – 5 = 18

Original equation.

| 2a – 3| – 5 = 18

2a – 3 = 23

Simplified inside absolute value.

| 2a – 3| = 23

2a = 26 a = 13

2a – 3 = – 23 2a = – 20 a = –10

5 added to both sides to isolate the absolute value.

Two equations formed because the expression inside the absolute value sign can equal +23 or – 23.

3 added to both sides.

Both sides divided by 2.

The numbers 13 and –10 will both satisfy the equation. This can be verified by substituting the solutions in the original equation.

Solve. 10.3

1. | 3x – 4 | = 5

4. | 2x – 10 | = 4

7. | – 3x – 3| = 24


2. | 4x – 2 | = 6

5. | 3x – 9 | = 24

8. | 12x – 4| = 2

3. | 5x – 5 | = 15

6. | 9x – 6 – 3x | – 8 = 46

9. | 8 + 4x | + 2 = 6

Review

Change from radical to exponential notation. Simplify if possible. 10.2 10. √e4

11. √7h

3

12. √k 4

5

13. √x 2y

Change from exponential to radical notation. Simplify if possible. 10.2 2

1

14. m 3

1

15. xy 2

16. 5 4

Simplify by rationalizing denominators. 9.7 √3 18. √10 19. 2 √7 √6 Simplify the radical expressions. 9.3 21.

√

60 9

22.

2

17. 9 3

20.

√27 + √300

23.

√3

2 √6x √2y

3

√24x 3 √3

3

Solve the quadratic equations for x by taking the square root of both sides. 9.2 24. 4x 2 = 64

25. x 2 = 8

26. x 2 – 14x + 49 = 36

Solve the quadratic equations by completing the square. 9.12 27. 2x 2 – 12x – 182 = 0

28. 2x 2 – 8x – 6 = 0

Solve by using the quadratic formula. 10.1 30. x 2 + 4x + 2 = 0

Simplify the expressions. 9.11 2 1 + k 3 33. k–1

Solve. 10.3

36. |2x – 3| = 4

31. 6x 2 – 5x + 1 = 0

34.

2y 3 3x 2z

5y 5 12xz 4

37. |6x + 5 – 3x| = 8

29. 2x 2 – 6x – 7 = 0

32. 16x 2 – 16x + 1 = 0

35.

2r + s 5s

6rs + 3s2

38. |4x – 5| + 7= 12

10.3 Absolute Value Equations ~ 11