8-64. Graph y = x 2 βˆ’ 2x – 15 without using the calculator. (Adjust

Solutions
Algebra: 8.2.2 Solving Quadratic Equations Name ______________________________
Block _____ Date ______
Bell Work 1/15 Factor each quadratic equation completely. Show sufficient work.
a. y = 2x2 + 8x + 8
2
𝑦 = 2(π‘₯ + 4π‘₯ + 4)
𝑦 = 𝟐( 𝒙 + 𝟐) 𝟐
2 2π‘₯
π‘₯ π‘₯2
π‘₯
4
b. y = 25x2 βˆ’ 64
𝑦 = (πŸ“π’™ + πŸ–)(πŸ“π’™ βˆ’ πŸ–)
2π‘₯
2
8-64. Graph y = x βˆ’ 2x – 15 without using the calculator. (Adjust scale as needed)
2
3 3π‘₯ βˆ’15
π‘₯ π‘₯
2
π‘₯
a. Factored form:
βˆ’5π‘₯
βˆ’5 b. Find the roots:
Explain how to find.
c. Find the y-intercept.
Explain how to find.
𝑦 = (π‘₯ + 3)(π‘₯ βˆ’ 5)
π‘₯+3=0
π‘₯βˆ’5=0
π‘₯ = βˆ’3
π‘₯=5
Set Factors equal to 0 and solve
-3
𝑦 = βˆ’15
It is the last number in
the standard equation
d. Find the x-coordinate of the vertex.
Explain how to find.
βˆ’3 + 5
=1
2
Average the roots
5
-15
(1,-16)
𝑦 = (1 + 3)(1 βˆ’ 5)
𝑦 = (4)(βˆ’4)
𝑦 = βˆ’16
Plug x into an equation and solve
2
8-65. Graph y = 2x + 5x – 12 without using the calculator.
e. Find the y-coordinate of the vertex.
Explain how to find.
4 8π‘₯ βˆ’12
π‘₯ 2π‘₯ 2 βˆ’3π‘₯ a. Factored form:
2π‘₯
βˆ’3 b. Find the roots:
π’š = (𝒙 + πŸ’)(𝟐𝟐 βˆ’ πŸ‘)
π‘₯+4 =0
𝒙 = βˆ’πŸ’
2π‘₯ βˆ’ 3 = 0
2π‘₯ = 3
𝒙 = 𝟏. πŸ“
c. Find the y-intercept.
π’š = βˆ’πŸπŸ
d. Find the x-coordinate of the vertex. βˆ’4 + 1.5
π‘₯=
= βˆ’1.25
2
e. Find the y-coordinate of the vertex.
𝑦 = (βˆ’1.25 + 4)(2(βˆ’1.25) βˆ’ 3)
𝑦 = (2.25)(βˆ’5.5) = βˆ’15.25
-4
1.5
-12
(-1.25, -15.25)
8-68. Graph y = x2 – 2x – 8 without using the calculator.
2 2π‘₯
βˆ’8
π‘₯ π‘₯ 2 βˆ’4π‘₯ a. Factored form:
π‘₯
βˆ’4 b. Find the roots:
π’š = (𝒙 + 𝟐)(𝒙 βˆ’ πŸ’)
𝒙 = βˆ’πŸ
c. Find the y-intercept. π’š = βˆ’πŸ–
𝒙=πŸ’
-2
d. Find the x-coordinate of the vertex. βˆ’πŸ + πŸ’
𝒙=
=𝟏
𝟐
e. Find the y-coordinate of the vertex.
π’š = 𝟏𝟐 βˆ’ 𝟐(𝟏) βˆ’ πŸ– = βˆ’πŸ—
4
-8
(1,-9)
8-69. Graph y = x2 + x – 6 without using the calculator.
3 3π‘₯
βˆ’6
π‘₯ π‘₯ 2 βˆ’2π‘₯ a.
π‘₯
Factored form:
βˆ’2 b.
Find the roots:
π’š = (𝒙 + πŸ‘)(𝒙 βˆ’ 𝟐)
𝒙 = βˆ’πŸ‘
𝒙=𝟐
π’š = βˆ’πŸ”
c.
Find the y-intercept.
d.
Find the x-coordinate of the vertex.
βˆ’πŸ‘ + 𝟐
𝒙=
= βˆ’πŸŽ. πŸ“
Find the y-coordinate of the vertex. 𝟐
e.
π’š = (βˆ’πŸŽ. πŸ“ + πŸ‘)(βˆ’πŸŽ. πŸ“ βˆ’ 𝟐)
π’š = (𝟐. πŸ“)(βˆ’πŸ. πŸ“)
π’š = βˆ’πŸ”. 𝟐𝟐
8-70. Compare the two equations below.
(x + 2)(x βˆ’ 1) = 0 and (x + 2) + (x βˆ’ 1) = 0
a. How are the equations different?
The first equation is multiplying
binomials (quadratic). The second
equation is adding binomials (linear)
b. Solve both equations.
(π‘₯ + 2)(π‘₯ βˆ’ 2) = 0
π‘₯+2 =0
π‘₯βˆ’1 =0
𝒙 = βˆ’πŸ
𝒙=𝟏
2 Solutions
π‘₯+2+π‘₯βˆ’1= 0
2π‘₯ + 1 = 0
2π‘₯ = βˆ’1
βˆ’πŸ
𝒙=
𝟐
1 Solution
The following are answers only. Show SUFFICIENT WORK!.
8-71. For each equation below, solve for x.
a.
(x βˆ’ 2)(x + 8) = 0
b.
(3x βˆ’ 9)(x βˆ’ 1) = 0
c.
(x + 10)(2x βˆ’ 5) = 0 𝒙 = βˆ’πŸπŸ
(x βˆ’ 7)2 = 0
d.
𝒙=𝟐
𝒙 = βˆ’πŸ–
𝒙=πŸ‘
𝒙=πŸ•
𝒙=𝟏
𝒙=
πŸ“
𝟐
8-72. Solve the system of equations.
5x βˆ’ 2y = 4
x=0
(𝟎, βˆ’πŸ)
8-73. The (x-intercepts) of the graph of y = 2x2 βˆ’ 16x + 30 are (3, 0) and (5, 0).
a.
b.
What is the x-coordinate of the vertex? How do you know?
4, it is halfway between 3 and 5.
Use your answer to part (a) above to find the y-coordinate of the vertex. Then
write the vertex as a point (x, y).
𝑦 = 2(4)2 βˆ’ 16(4) + 30 = βˆ’2
8-74. Factor each quadratic below completely.
a.
2x2 βˆ’ 2x βˆ’ 4
b.
4x2 βˆ’ 24x + 36 = πŸ’(𝒙 βˆ’ πŸ‘)𝟐
= 𝟐(𝒙 βˆ’ 𝟐)(𝒙 + 𝟏)
EOC Practice:
βˆ’πŸ’
πŸ’
𝟏𝟏
𝟏𝟏
(πŸ’, βˆ’πŸ)