1. On Page 234, complete the Prerequisite skills #1-14.

2. Chapter 4: Quadratic Function and Factoring
BIG IDEAS:
Graphing and writing quadratic functions in several forms
Solving quadratic equations using a variety of methods
Performing operations with square roots and complex numbers

3. Lesson 1: Graph Quadratic Functions in Standard Form

4. Essential question
How are the values of a, b, and c in the equation y = ax2 + bx + c related to the graph of a quadratic function?

5. VOCABULARY Quadratic Function – y = ax2 + bx + c
Parabola – The set of all points equidistant from a point called the focus and a line called the directrix.
Vertex – The point on a parabola that lies on the axis of symmetry
Axis of Symmetry – the line perpendicular to the parabola’s directrix and passing through its focus and vertex

6. EXAMPLE 1
Graph a function of the form y = ax2
STEP 4
Compare the graphs of y = 2x2 and y = x2.
Both open up and have the same vertex and
axis of symmetry. The graph of y = 2x2 is
narrower than the graph of y = x2.

7. EXAMPLE 2
Graph a function of the form y = ax2 + c
x2 + 3 12
Graph y = – Compare the graph with the
graph of y = x2
SOLUTION
Make a table of values for y = –
x2 + 3 12
STEP 1
STEP 2
Plot the points from the table.
STEP 3
Draw a smooth curve through the points.

8. EXAMPLE 2
Graph a function of the form y = ax2
x2 + 3 12
STEP 4
Compare the graphs of y = –
and y = x2.
Both graphs have the same axis of symmetry. However, the graph of y = – opens down and is wider than the graph of y = x2. Also, its vertex is 3 units higher.
x2 + 3 12

9. GUIDED PRACTICE
for Examples 1 and 2
Graph the function. Compare the graph with the graph of y = x2.
1. y = – 4x2
ANSWER
Same axis of symmetry and vertex, opens down, and is narrower

10. GUIDED PRACTICE
for Examples 1 and 2
2. y = – x2 – 5
ANSWER
Same axis of symmetry, vertex is shifted down 5 units, and opens down

11. GUIDED PRACTICE
for Examples 1 and 2
x2 + 2 14
3. f(x) =
ANSWER
Same axis of symmetry, vertex is shifted up 2 units, opens up, and is wider

12. b 2a (–8) 2(2) EXAMPLE 3 Graph a function of the form y = ax2 + bx + c
Graph y = 2x2 – 8x + 6.
SOLUTION
STEP 1
Identify the coefficients of the function. The coefficients are a = 2, b = –8, and c = 6. Because a > 0, the parabola opens up.
STEP 2
Find the vertex. Calculate the x-coordinate.
x =
b 2a =
(–8) (2) –
= 2
Then find the y-coordinate of the vertex.
y = 2(2)2 – 8(2) + 6 = –2
So, the vertex is (2, –2). Plot this point.

13. EXAMPLE 3
Graph a function of the form y = ax2 + bx + c
STEP 3
Draw the axis of symmetry x = 2.
STEP 4
Identify the y-intercept c, which is 6. Plot the point (0, 6). Then reflect this point in the axis of symmetry to plot another point, (4, 6).
STEP 5
Evaluate the function for another value of x, such as x = 1.
y = 2(1)2 – 8(1) + 6 = 0
Plot the point (1,0) and its reflection (3,0) in the axis of symmetry.

14. EXAMPLE 3
Graph a function of the form y = ax2 + bx + c
STEP 6
Draw a parabola through the plotted points.

15. GUIDED PRACTICE
for Example 3
Graph the function. Label the vertex and axis of symmetry.
4. y = x2 – 2x – 1
5. y = 2x2 + 6x + 3

16. GUIDED PRACTICE
for Example 3
6. f (x) = x2 – 5x + 2
1 3 –

17. A – tell whether the parabola opens up or down
Essential question
How are the values of a, b, and c in the equation y = ax2 + bx + c related to the graph of a quadratic function?
A – tell whether the parabola opens up or down
A/b – used to find the equation of the axis of symmetry and x-coordinate of the vertex
C – y-intercept

18. Find the product a. (x + 6) (x + 3) b. (x-5)2 c. 4(x+5)(x-5)

19. Lesson 2: Graph Quadratic Functions in Vertex or intercept form

20. Essential question
Why do we write the graph of quadratic functions in vertex form or intercept form?

21. VOCABULARY Vertex form – y = a(x – h)2 + k
Intercept form – y = a(x-p)(x-q)

22. EXAMPLE 1
Graph a quadratic function in vertex form 14
Graph y = – (x + 2)2 + 5.
SOLUTION
STEP 1
Identify the constants a = – , h = – 2, and k = 5.
Because a < 0, the parabola opens down. 14
STEP 2
Plot the vertex (h, k) = (– 2, 5) and draw the axis of symmetry x = – 2.

23. EXAMPLE 1
Graph a quadratic function in vertex form
STEP 3
Evaluate the function for two values of x.
x = 0: y = (0 + 2)2 + 5 = 4 14 –
x = 2: y = (2 + 2)2 + 5 = 1 14 –
Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry.
STEP 4
Draw a parabola through the plotted points.

24. EXAMPLE 2
Use a quadratic model in vertex form
Civil Engineering
The Tacoma Narrows Bridge in Washington has two towers that each rise 307 feet above the roadway and are connected by suspension cables as shown. Each cable can be modeled by the function.
y = (x – 1400)2 + 27
1 7000
where x and y are measured in feet. What is the distance d between the two towers ?

25. EXAMPLE 2
Use a quadratic model in vertex form
SOLUTION
The vertex of the parabola is (1400, 27). So, a cable’s lowest point is 1400 feet from the left tower shown above. Because the heights of the two towers are the same, the symmetry of the parabola implies that the vertex is also 1400 feet from the right tower. So, the distance between the two towers is d = 2 (1400) = 2800 feet.

26. GUIDED PRACTICE
for Examples 1 and 2
Graph the function. Label the vertex and axis of symmetry.
1. y = (x + 2)2 – 3
2. y = –(x + 1)2 + 5

27. GUIDED PRACTICE
for Examples 1 and 2 12
3. f(x) = (x – 3)2 – 4

28. GUIDED PRACTICE
for Examples 1 and 2
4. WHAT IF? Suppose an architect designs a bridge
y = 1
6500
(x – 1400)
with cables that can be modeled by
where x and y are measured in feet. Compare this function’s graph to the graph of the function in Example 2.
ANSWER
This graph is slightly steeper than the graph in Example 2. They both have the same vertex and axis of symmetry, and both open up.

29. EXAMPLE 3
Graph a quadratic function in intercept form
Graph y = 2(x + 3)(x – 1).
SOLUTION
STEP 1
Identify the x-intercepts. Because p = –3 and q = 1, the x-intercepts occur at the points (–3, 0) and (1, 0).
STEP 2
Find the coordinates of the vertex.
x =
p + q 2
–3 + 1
= –1 =
y = 2(–1 + 3)(–1 – 1) = –8
So, the vertex is (–1, –8)

30. EXAMPLE 3
Graph a quadratic function in intercept form
STEP 3
Draw a parabola through the vertex and the points where the x-intercepts occur.

31. EXAMPLE 4
Use a quadratic function in intercept form
Football
The path of a placekicked football can be modeled by the function y = –0.026x(x – 46) where x is the horizontal distance (in yards) and y is the corresponding height (in yards).
a. How far is the football kicked ?
b. What is the football’s maximum height ?

32. EXAMPLE 4
Use a quadratic function in intercept form
SOLUTION
a. Rewrite the function as y = –0.026(x – 0)(x – 46). Because p = 0 and q = 46, you know the x-intercepts are 0 and 46. So, you can conclude that the football is kicked a distance of 46 yards.
b. To find the football’s maximum height, calculate the coordinates of the vertex.
x =
p + q 2
0 + 46
= 23 =
y = –0.026(23)(23 – 46)
The maximum height is the y-coordinate of the vertex, or about 13.8 yards.

33. GUIDED PRACTICE
for Examples 3 and 4
Graph the function. Label the vertex, axis of symmetry, and x-intercepts.
5. y = (x – 3)(x – 7)
f (x) = 2(x – 4)(x + 1)

34. GUIDED PRACTICE
for Examples 3 and 4
y = –(x + 1)(x – 5)
8. WHAT IF? In Example 4, what is the maximum height of the football if the football’s path can be modeled by the function y = –0.025x(x – 50)?
yards
ANSWER

35. Write original function. Multiply using FOIL.
EXAMPLE 5
Change from intercept form to standard form
Write y = –2(x + 5)(x – 8) in standard form.
y = –2(x + 5)(x – 8)
Write original function.
= –2(x2 – 8x + 5x – 40)
Multiply using FOIL.
= –2(x2 – 3x – 40)
Combine like terms.
= –2x2 + 6x + 80
Distributive property

36. Write original function. Rewrite (x – 1)2.
EXAMPLE 6
Change from vertex form to standard form
Write f (x) = 4(x – 1)2 + 9 in standard form.
f (x) = 4(x – 1)2 + 9
Write original function.
= 4(x – 1) (x – 1) + 9
Rewrite (x – 1)2.
= 4(x2 – x – x + 1) + 9
Multiply using FOIL.
= 4(x2 – 2x + 1) + 9
Combine like terms.
= 4x2 – 8x
Distributive property
= 4x2 – 8x + 13
Combine like terms.

37. GUIDED PRACTICE
for Examples 5 and 6
Write the quadratic function in standard form.
9. y = –(x – 2)(x – 7)
f(x) = 2(x + 5)(x + 4)
ANSWER
ANSWER
–x2 + 9x – 14
2x2 + 18x + 40
y = – 4(x – 1)(x + 3)
y = –7(x – 6)(x + 1)
ANSWER
ANSWER
–4x2 – 8x + 12
–7x2 + 35x + 42

38. GUIDED PRACTICE
for Examples 5 and 6
13. y = –3(x + 5)2 – 1
f(x) = –(x + 2)2 + 4
ANSWER
ANSWER
–3x2 – 30x – 76
–x2 – 4x
g(x) = 6(x – 4)2 – 10
16. y = 2(x – 3)2 + 9
ANSWER
ANSWER
6x2 – 48x + 86
2x2 – 12x + 27

39. Essential question
Why do we write the graph of quadratic functions in vertex form or intercept form?
When a quadratic function is written in vertex form, you can read the coordinates of the vertex directly from the equation. When a quadratic function is written in intercept form, you can read the x-intercepts directly from the equation.

40. Find the product: a. (m-8) (M-9) b. (d+9)2 c. (y+20) (y-20)

41. Lesson 3: Solve x2 + bx + c = 0 by Factoring

42. How can factoring be used to solve quadratic equations when A = 1?
Essential question
How can factoring be used to solve quadratic equations when A = 1?

43. VOCABULARY
Monomial – An expression that is either a number, variable or the product of a number and one or more variables with whole number exponents
Binomial – The sum of two monomials
Trinomial – The sum of three monomials
Quadratic Equation – ax2 + bx + c = 0, a≠ 0
Root of an equation – The solution of a quadratic equation

44. EXAMPLE 1
Factor trinomials of the form x2 + bx + c
Factor the expression.
a. x2 – 9x + 20
b. x2 + 3x – 12
SOLUTION
a. You want x2 – 9x + 20 = (x + m)(x + n) where mn = 20 and m + n = –9.
ANSWER
Notice that m = –4 and n = –5.
So, x2 – 9x + 20 = (x – 4)(x – 5).

45. EXAMPLE 1
Factor trinomials of the form x2 + bx + c
b. You want x2 + 3x – 12 = (x + m)(x + n) where mn = – 12 and m + n = 3.
ANSWER
Notice that there are no factors m and n such that m + n = 3. So, x2 + 3x – 12 cannot be factored.

46. GUIDED PRACTICE
for Example 1
Factor the expression. If the expression cannot be factored, say so.
1. x2 – 3x – 18
2. n2 – 3n + 9
3. r2 + 2r – 63
ANSWER
ANSWER
ANSWER
(x – 6)(x + 3)
cannot be factored
(r + 9)(r –7)

47. Difference of two squares
EXAMPLE 2
Factor with special patterns
Factor the expression.
a. x2 – 49
= x2 – 72
Difference of two squares
= (x + 7)(x – 7)
b. d d + 36
= d 2 + 2(d)(6) + 62
Perfect square trinomial
= (d + 6)2
c. z2 – 26z + 169
= z2 – 2(z) (13) + 132
Perfect square trinomial
= (z – 13)2

48. GUIDED PRACTICE
for Example 2
Factor the expression.
4. x2 – 9
ANSWER
(x – 3)(x + 3)
5. q2 – 100
ANSWER
(q – 10)(q + 10)
6. y2 + 16y + 64
ANSWER
(y + 8)2

49. GUIDED PRACTICE
for Example 2
7. w2 – 18w + 81
(w – 9)2

50. How can factoring be used to solve quadratic equations when a = 1?
Essential question
How can factoring be used to solve quadratic equations when a = 1?
The left side of ax2 + bx + c = 0 can be factored, factor the trinomial, use the zero product property to set each factor equal to 0 and solve resulting linear equations.

51. Find the product: a. (4y-3)(3y+8) b. (5m+6)(5m-6)

52. Lesson 4: Solve Ax2 + bx + c = 0 by Factoring

53. How can factoring be used to solve quadratic equations when
Essential question
How can factoring be used to solve quadratic equations when
a ≠ 1?

54. VOCABULARY
There is no new vocabulary for this lesson.

55. Difference of two squares
EXAMPLE 3
Factor with special patterns
Factor the expression.
a. 9x2 – 64
= (3x)2 – 82
Difference of two squares
= (3x + 8)(3x – 8)
b. 4y2 + 20y + 25
= (2y)2 + 2(2y)(5) + 52
Perfect square trinomial
= (2y + 5)2
c w2 – 12w + 1
= (6w)2 – 2(6w)(1) + (1)2
Perfect square trinomial
= (6w – 1)2

56. GUIDED PRACTICE
GUIDED PRACTICE
for Example 3
Factor the expression.
x2 – 1
ANSWER
(4x + 1)(4x – 1)
y2 + 12y + 4
ANSWER
(3y + 2)2
r2 – 28r + 49
ANSWER
(2r – 7)2
s2 – 80s + 64
ANSWER
(5s – 8)2

57. GUIDED PRACTICE
GUIDED PRACTICE
for Example 3
z2 + 4z + 9
ANSWER
(7z + 3)2
n2 – 9 = (3y)2
ANSWER
(6n – 3)(6n +3)

58. EXAMPLE 4
Factor out monomials first
Factor the expression.
a. 5x2 – 45
= 5(x2 – 9)
= 5(x + 3)(x – 3)
b. 6q2 – 14q + 8
= 2(3q2 – 7q + 4)
= 2(3q – 4)(q – 1)
c. –5z2 + 20z
= –5z(z – 4)
d. 12p2 – 21p + 3
= 3(4p2 – 7p + 1)

59. GUIDED PRACTICE
GUIDED PRACTICE
for Example 4
Factor the expression.
s2 – 24
ANSWER
3(s2 – 8)
t2 + 38t – 10
ANSWER
2(4t – 1) (t + 5)
x2 + 24x + 15
ANSWER
3(2x2 + 8x + 5)
x2 – 28x – 24
ANSWER
4(3x + 2)(x – 3)
–16n2 + 12n
ANSWER
–4n(4n – 3)

60. GUIDED PRACTICE
GUIDED PRACTICE
for Example 4
z2 + 33z + 36
ANSWER
3(2z + 3)(z + 4)

61. Write original equation.
EXAMPLE 5
Solve quadratic equations
Solve (a) 3x2 + 10x – 8 = 0 and (b) 5p2 – 16p + 15 = 4p – 5.
a. 3x2 + 10x – 8 = 0
Write original equation.
(3x – 2)(x + 4) = 0
Factor.
3x – 2 = 0
or x + 4 = 0
Zero product property
or x = –4
x = 23
Solve for x.

62. Write original equation.
EXAMPLE 5
Solve quadratic equations
b. 5p2 – 16p + 15 = 4p – 5.
Write original equation.
5p2 – 20p + 20 = 0
Write in standard form.
p2 – 4p + 4 = 0
Divide each side by 5.
(p – 2)2 = 0
Factor.
p – 2 = 0
Zero product property
p = 2
Solve for p.

63. Essential question
How can factoring be used to solve quadratic equations when a ≠ 1?
First, write the equation in standard form.
Then factor any common monomial.
Next, factor the expression.
Use the zero product property to solve the equation.

64. Find the exact value: a. 𝟒𝟗 b. - 𝟏𝟒𝟒

65. Lesson 5: Solve quadratic equations by finding square roots

66. How can you use square roots to solve a quadratic equation?
Essential question
How can you use square roots to solve a quadratic equation?

67. VOCABULARY
Square root – A number that is multiplied by itself to solve a square
Radical – An expression of the form 𝑥 or 𝑛 𝑥 where x is a number or an expression
Radicand – The number or expression beneath a radical sign
Conjugate – The expression a + 𝑏 and a - 𝑏 where and a b are rational numbers.

68. EXAMPLE 1
Use properties of square roots
Simplify the expression. a. 80 5 16 = 5
= 4 b. 6 21
126 = 9 14 =
= 3 14 c. 4 81 = 4 81 =
2 9 d. 7 16 = 7 16 = 4 7

69. GUIDED PRACTICE
GUIDED PRACTICE
for Example 1 27
ANSWER 3 98
ANSWER 2 7

70. GUIDED PRACTICE
GUIDED PRACTICE
for Example 1 10 15
ANSWER 6 5 8 28
ANSWER 14 4

71. GUIDED PRACTICE
GUIDED PRACTICE
for Example 1 9 64
3 8
ANSWER 15 4 2 15
ANSWER

72. GUIDED PRACTICE
GUIDED PRACTICE
for Example 1 11 25 5 11
ANSWER 36 49 7 6
ANSWER

73. EXAMPLE 2
Rationalize denominators of fractions. 5 2 3
7 + 2
Simplify
(a)
and
(b)
SOLUTION
(a) 5 2 = 5 2 = 5 2 2 10 =

74. EXAMPLE 2
Rationalize denominators of fractions.
SOLUTION = 3
7 + 2
7 –
(b) 3
7 + 2 =
21 – 3 2
49 – – 2 =
21 – 3 2 47

75. Write original equation.
EXAMPLE 3
Solve a quadratic equation
Solve 3x2 + 5 = 41.
3x2 + 5 = 41
Write original equation.
3x2 = 36
Subtract 5 from each side.
x2 = 12
Divide each side by 3.
x = + 12
Take square roots of each side.
x = + 4 3
Product property
x =
+ 2 3
Simplify.

76. Solve a quadratic equation
EXAMPLE 3
Solve a quadratic equation
ANSWER
The solutions are and 2 3 2 3 –
Check the solutions by substituting them into the original equation.
3x2 + 5 = 41
3x2 + 5 = 41
3( )2 + 5 = 41 2 3 ?
3( )2 + 5 = 41
– 2 3 ?
3(12) + 5 = 41 ?
3(12) + 5 = 41 ?
41 = 41 
41 = 41 

77. Write original equation.
EXAMPLE 4
Standardized Test Practice
SOLUTION 15
(z + 3)2 = 7
Write original equation.
(z + 3)2 = 35
Multiply each side by 5.
z + 3 = + 35
Take square roots of each side.
z = –3 + 35
Subtract 3 from each side.
The solutions are – and –3 – 35

78. EXAMPLE 4
Standardized Test Practice
ANSWER
The correct answer is C.

79. GUIDED PRACTICE
GUIDED PRACTICE
for Examples 2, 3, and 4
Simplify the expression. 6 5 19 21 5 30
399 21
ANSWER
ANSWER 9 8
– 6
7 – 5 2 4 3
– 21 – 3 5 22
ANSWER
ANSWER 17 12 2
4 + 11 51 6
ANSWER
8 – 2 11 5
ANSWER

80. GUIDED PRACTICE
for Examples 2, 3, and 4
– 1
9 + 7
– 9 + 7 74
ANSWER 4
8 – 3
32 + 4 3 61
ANSWER

81. GUIDED PRACTICE
for Examples 2, 3, and 4
Solve the equation.
5x2 = 80
ANSWER
+ 4
z2 – 7 = 29
+ 6
ANSWER
3(x – 2)2 = 40
120 3 2 +
ANSWER

82. How can you use square roots to solve a quadratic equation?
Essential question
How can you use square roots to solve a quadratic equation?
If a quadratic equation is written in the form x2 = s where s > 0, then you can solve it by taking the square roots of both sides and simplifying the results.

83. Solve the equation: A. 3x2 + 8 = 23 b. 2(x+7)2 = 16

84. Lesson 6: perform operations with complex numbers

85. How do you perform opations on complex numbers?
Essential question
How do you perform opations on complex numbers?

86. VOCABULARY Imaginary unit (i) – i = −1 so i2
Complex number – A number a + bi where a and b are real numbers and i is the imaginary unit

87. Write original equation.
EXAMPLE 1
Solve a quadratic equation
Solve 2x = –37.
2x = –37
Write original equation.
2x2 = –48
Subtract 11 from each side.
x2 = –24
Divide each side by 2.
x = + –24
Take square roots of each side.
x = + i 24
Write in terms of i.
x = + 2i 6
Simplify radical.
ANSWER
The solutions are 2i and –2i

88. GUIDED PRACTICE
for Example 1
Solve the equation. 1.
x2 = –13. 4.
x2 – 8 = –36 .
ANSWER
+ i 13
ANSWER
+ 2i 7 2.
x2 = –38. 5.
3x2 – 7 = –31 .
ANSWER
+ i 38
ANSWER
+ 2i 3.
x2 + 11= 3. 6.
5x = 3 .
ANSWER
+ 2i 2
ANSWER
+ i 6

89. Definition of complex addition
EXAMPLE 2
Add and subtract complex numbers
Write the expression as a complex number in standard form.
a. (8 – i) + (5 + 4i)
b. (7 – 6i) – (3 – 6i)
c. 10 – (6 + 7i) + 4i
SOLUTION
a. (8 – i) + (5 + 4i) =
Definition of complex addition
(8 + 5) + (–1 + 4)i
= i
Write in standard form.
b. (7 – 6i) – (3 – 6i) =
Definition of complex subtraction
(7 – 3) + (–6 + 6)i
= 4 + 0i
Simplify.
= 4
Write in standard form.

90. Definition of complex subtraction
EXAMPLE 2
Add and subtract complex numbers
c. 10 – (6 + 7i) + 4i =
Definition of complex subtraction
[(10 – 6) – 7i] + 4i
= (4 – 7i) + 4i
Simplify.
= 4 + (–7 + 4)i
Definition of complex addition
= 4 – 3i
Write in standard form.

91. GUIDED PRACTICE
for Example 2
Write the expression as a complex number in standard form.
7. (9 – i) + (–6 + 7i)
ANSWER
3 + 6i
8. (3 + 7i) – (8 – 2i)
ANSWER
–5 + 9i
9. –4 – (1 + i) – (5 + 9i)
ANSWER
–10 – 10i

92. Distributive property
EXAMPLE 4
Multiply complex numbers
Write the expression as a complex number in standard form.
a. 4i(–6 + i)
b. (9 – 2i)(–4 + 7i)
SOLUTION
a. 4i(–6 + i) = –24i + 4i2
Distributive property
= –24i + 4(–1)
Use i2 = –1.
= –24i – 4
Simplify.
= –4 – 24i
Write in standard form.

93. Multiply using FOIL. Simplify and use i2 = – 1 . Simplify.
EXAMPLE 4
Multiply complex numbers
b. (9 – 2i)(–4 + 7i)
= – i + 8i – 14i2
Multiply using FOIL.
= – i – 14(–1)
Simplify and use i2 = – 1 .
= – i + 14
Simplify.
= – i
Write in standard form.

94. EXAMPLE 5
Divide complex numbers
Write the quotient in standard form.
7 + 5i
1  4i
7 + 5i
1 – 4i =
1 + 4i
Multiply numerator and denominator by 1 + 4i, the complex conjugate of 1 – 4i.
7 + 28i + 5i + 20i2
1 + 4i – 4i – 16i2 =
Multiply using FOIL.
7 + 33i + 20(–1)
1 – 16(–1) =
Simplify and use i2 = 1.
– i 17 =
Simplify.

95. Write in standard form. EXAMPLE 5 Divide complex numbers 13 17 – = +33 i
Write in standard form.

96. GUIDED PRACTICE
for Examples 3, 4 and 5
10.
WHAT IF? In Example 3, what is the impedance of the circuit if the given capacitor is replaced with one having a reactance of 7 ohms?
5 – 4i ohms.
ANSWER

97. GUIDED PRACTICE
for Examples 3, 4 and 5
Write the expression as a complex number in standard form. 5
1 + i
11.
i(9 – i)
13. 5 2 – i
ANSWER
1 + 9i
ANSWER
14.
5 + 2i
12.
(3 + i)(5 – i)
3 – 2i 11 13 + 16 i
ANSWER
16 + 2i
ANSWER

98. Essential question How do you perform operations on complex numbers?
Add/subtract – add or subtract their real parts and their imaginary parts separately.
Multiply – use the distributive property or the FOIL method.
DIVIDE – multiply the numerator and denominator by the complex conjugate of the denominator

99. Factor the expression: a. x2 + 18x + 81 b. x2 – 22x + 121

100. Lesson 7: Complete the square

101. Essential question
How is the process of completing the square used to solve quadratic equations?

102. VOCABULARY
Completing the square – The process of adding a term to a quadratic expression of the form x2 + bx to make it a perfect square trinomial

103. Write original equation.
EXAMPLE 1
Solve a quadratic equation by finding square roots
Solve x2 – 8x + 16 = 25.
x2 – 8x + 16 = 25
Write original equation.
(x – 4)2 = 25
Write left side as a binomial squared.
x – 4 = +5
Take square roots of each side.
x = 4 + 5
Solve for x.
The solutions are = 9 and 4 –5 = – 1.
ANSWER

104. EXAMPLE 2
Make a perfect square trinomial
Find the value of c that makes x2 + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial.
SOLUTION
STEP 1 16 2 = 8
Find half the coefficient of x.
STEP 2
Square the result of Step 1.
82 = 64
STEP 3
Replace c with the result of Step 2.
x2 + 16x + 64

105. EXAMPLE 2
Make a perfect square trinomial
ANSWER
The trinomial x2 + 16x + c is a perfect square when c = 64. Then x2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.

106. GUIDED PRACTICE
for Examples 1 and 2
Solve the equation by finding square roots.
x2 + 6x + 9 = 36.
ANSWER
3 and –9.
x2 – 10x + 25 = 1.
ANSWER
4 and 6.
x2 – 24x = 100.
ANSWER
2 and 22.

107. GUIDED PRACTICE
for Examples 1 and 2
Find the value of c that makes the expression a perfect square trinomial.Then write the expression as the square of a binomial. 4.
x2 + 14x + c
ANSWER
49 ; (x + 7)2 5.
x2 + 22x + c
ANSWER
121 ; (x + 11)2 6.
x2 – 9x + c 81 4 9 2
ANSWER
; (x – )2.

108. ( ) Write original equation. Write left side in the form x2 + bx. Add
EXAMPLE 3
Solve ax2 + bx + c = 0 when a = 1
Solve x2 – 12x + 4 = 0 by completing the square.
x2 – 12x + 4 = 0
Write original equation.
x2 – 12x = –4
Write left side in the form x2 + bx.
x2 – 12x + 36 = –4 + 36
Add
–12 2 ( ) =
(–6) 36
to each side.
(x – 6)2 = 32
Write left side as a binomial squared.
x – 6 =
Take square roots of each side.
x =
Solve for x.
x =
Simplify: 32 = 16 2 4
The solutions are 6 + 4
and 6 – 4 2
ANSWER

109. EXAMPLE 3
Solve ax2 + bx + c = 0 when a = 1
CHECK
You can use algebra or a graph.
Algebra Substitute each solution in the original equation to verify that it is correct.
Graph Use a graphing calculator to graph
y = x2 – 12x + 4. The x-intercepts are about – 4 2 and

110. ( ) Write original equation.
EXAMPLE 4
Solve ax2 + bx + c = 0 when a = 1
Solve 2x2 + 8x + 14 = 0 by completing the square.
2x2 + 8x + 14 = 0
Write original equation.
x2 + 4x + 7 = 0
Divide each side by the coefficient of x2.
x2 + 4x = –7
Write left side in the form x2 + bx.
Add 4 2 ( ) =
to each side.
x2 – 4x + 4 = –7 + 4
(x + 2)2 = –3
Write left side as a binomial squared.
x + 2 = –3
Take square roots of each side.
x = – –3
Solve for x.
x = –2 + i 3
Write in terms of the imaginary unit i.

111. EXAMPLE 4
Solve ax2 + bx + c = 0 when a = 1
The solutions are –2 + i 3
and –2 – i
3 .
ANSWER

112. EXAMPLE 5
Standardized Test Practice
SOLUTION
Use the formula for the area of a rectangle to write an equation.

113. ( ) Length Width = Area Distributive property
EXAMPLE 5
Standardized Test Practice
3x(x + 2) = 72
Length Width = Area
3x2 + 6x = 72
Distributive property
x2 + 2x = 24
Divide each side by the coefficient of x2.
Add 2 ( ) = 1
to each side.
x2 – 2x + 1 =
(x + 1)2 = 25
Write left side as a binomial squared.
x + 1 = + 5
Take square roots of each side.
x = –1 + 5
Solve for x.

114. EXAMPLE 5
Standardized Test Practice
So, x = –1 + 5 = 4 or x = – 1 – 5 = –6. You can reject x = –6 because the side lengths would be –18 and –4, and side lengths cannot be negative.
The value of x is 4. The correct answer is B.
ANSWER

115. GUIDED PRACTICE
for Examples 3, 4 and 5
Solve the equation by completing the square. 7.
x2 + 6x + 4 = 0
10.
3x2 + 12x – 18 = 0
–3+ 5
ANSWER
–2 + 10
ANSWER 8.
x2 – 10x + 8 = 0
11.
6x(x + 8) = 12
5 + 17
ANSWER
–4 +3 2
ANSWER 9.
2n2 – 4n – 14 = 0
12.
4p(p – 2) = 100
1 + 2 2
ANSWER
1 + 26
ANSWER

116. Essential question
How is the process of completing the square used to solve quadratic equations?
You complete the square so that one side of the equation can be written as the square of a binomial. Then you take the square roots of both sides and simplify the results.

117. Evaluate b2 – 4ac when a = 3, b = -6 and c = 5.

118. Lesson 8: use the quadratic formula and the discriminant

119. How do you use the quadratic formula and the discriminant?
Essential question
How do you use the quadratic formula and the discriminant?

120. VOCABULARY
Quadratic formula – The formula x = −𝑏 ± − 𝑏 2 −4𝑎𝑐 2𝑎 used to find the solution of the quadratic equation ax2 + bx + c = 0
Discriminant – The expression b2 + bx + c = 0; also the expression under the radical sign of the quadratic formula

121. EXAMPLE 4
Use the discriminant
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
a. x2 – 8x + 17 = 0
b. x2 – 8x + 16 = 0
c. x2 – 8x + 15 = 0
SOLUTION
Equation
Discriminant
Solution(s)
ax2 + bx + c = 0
b2 – 4ac
x =
– b+ b2– 4ac
2ac
a. x2 – 8x + 17 = 0
(–8)2 – 4(1)(17) = –4
Two imaginary: 4 + i
b. x2 – 8x + 16 = 0
(–8)2 – 4(1)(16) = 0
One real: 4
b. x2 – 8x + 15 = 0
(–8)2 – 4(1)(15) = 0
Two real: 3,5

122. GUIDED PRACTICE
for Example 4
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
2x2 + 4x – 4 = 0
ANSWER
48 ; Two real solutions
3x2 + 12x + 12 = 0
ANSWER
0 ; One real solution
8x2 = 9x – 11
ANSWER
–271 ; Two imaginary solutions

123. GUIDED PRACTICE
for Example 4
7x2 – 2x = 5
ANSWER
144 ; Two real solutions
4x2 + 3x + 12 = 3 – 3x
ANSWER
–108 ; Two imaginary solutions
3x – 5x2 + 1 = 6 – 7x
ANSWER
0 ; One real solution

124. How do you use the quadratic formula and the discriminant?
Essential question
How do you use the quadratic formula and the discriminant?
Substitute the three coefficients from the standard form into the formula and simplify the result. The discriminant gives the number and type of solutions of a quadratic equation.