This document discusses solving polynomial equations by factoring. It provides an example of solving the equation x^2 - 2x - 3 = 0 by: 1) factoring the trinomial to get (x - 3)(x + 1) = 0, 2) setting each factor equal to 0 to get x - 3 = 0 and x + 1 = 0, and 3) extracting the solutions x = 3 and x = -1. It then works through another example of solving 2 = 2x^2 - 3x by similar factoring steps to get the solutions x = -1/2, x = 2. Finally, it introduces word problems that can be modeled by quadratic equations of the form AB = C.

1. Solving Equations by Factoring

2. Example A. a. Solve x2 – 2x – 3 = 0
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring

3. Example A. a. Solve x2 – 2x – 3 = 0
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,

4. Example A. a. Solve x2 – 2x – 3 = 0
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
One side is already 0.

5. Example A. a. Solve x2 – 2x – 3 = 0
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
One side is already 0.

6. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring

7. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring

8. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0 There are two linear factors.
We may extract one answer
from each.
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring

9. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
Hence x – 3 = 0
x = 3
There are two linear factors.
We may extract one answer
from each.
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring

10. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3 or x = –1
There are two linear factors.
We may extract one answer
from each.
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring

11. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3 or x = –1
There are two linear factors.
We may extract one answer
from each.
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring
b. Solve 2 = 2x2 – 3x

12. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3 or x = –1
There are two linear factors.
We may extract one answer
from each.
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring
b. Solve 2 = 2x2 – 3x
Make the left side 0 by moving the 2 so 0 = 2x2 – 3x – 2

13. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3 or x = –1
There are two linear factors.
We may extract one answer
from each.
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring
b. Solve 2 = 2x2 – 3x
Make the left side 0 by moving the 2 so 0 = 2x2 – 3x – 2
Factoring the trinomial yield 0 = (2x + 1)(x – 2),

14. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3 or x = –1
There are two linear factors.
We may extract one answer
from each.
An important factoring application is to find solutions of higher
degree equations via the following steps.
Solving Equations by Factoring
b. Solve 2 = 2x2 – 3x
Make the left side 0 by moving the 2 so 0 = 2x2 – 3x – 2
Factoring the trinomial yield 0 = (2x + 1)(x – 2),
hence 2x + 1 = 0 so x = – ½, or x – 2 = 0 or x = 2.

15. Example A. a. Solve x2 – 2x – 3 = 0
1. To solve a polynomial equation, set one side of the
equation to be 0 by moving all the terms the other side,
2. factor the polynomial, if possible,
3. then extract an answer from each binomial factor.
One side is already 0. Factoring the trinomial we have that
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3 or x = –1
There are two linear factors.
We may extract one answer
from each.
An important factoring application is to find solutions of higher
degree equations via the following steps.
Following are some applications of 2nd degree equations.
Solving Equations by Factoring
b. Solve 2 = 2x2 – 3x
Make the left side 0 by moving the 2 so 0 = 2x2 – 3x – 2
Factoring the trinomial yield 0 = (2x + 1)(x – 2),
hence 2x + 1 = 0 so x = – ½, or x – 2 = 0 or x = 2.

16. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
2nd-Degree-Equation Word Problems

17. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
2nd-Degree-Equation Word Problems

18. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
2nd-Degree-Equation Word Problems

19. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product C
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.

20. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product C
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x.

21. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product C
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).

22. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product C
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
The product of them is 96 so
x(3x – 2) = 96

23. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product C
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
The product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96

24. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product C
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
The product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0
make the right side 0

25. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product C
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
The product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0
(3x + 16)(x – 6) = 0
So x = –16/3 or x = 6.
make the right side 0

26. The simplest type of formulas leads to 2nd degree equations
are the ones of the form AB = C.
Specifically, if the two factors A and B are linear binomials
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product C
is known, is a 2nd degree equation.
2nd-Degree-Equation Word Problems
Example A. We have two positive numbers.
The larger number is 2 less than three times of the smaller one.
The product of two numbers is 96. Find the two numbers.
Let the smaller number be x. Hence the larger one is (3x – 2).
The product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0.
(3x + 16)(x – 6) = 0
So x = –16/3 or x = 6. Hence the two numbers are 6 and 16.
make the right side 0

27. Many physics formulas are 2nd degree.
2nd-Degree-Equation Word Problems

28. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems

29. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
where
h = height in feet
t = time in second
v = upward speed in feet per second

30. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second

31. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?

32. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64,

33. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64, so
h = -16(1)2 + 64(1)

34. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64, so
h = -16(1)2 + 64(1)
h = -16 + 64 = 48 ft

35. Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
2nd-Degree-Equation Word Problems
height = -16t2 + vt
after t secondswhere
h = height in feet
t = time in second
v = upward speed in feet per second
Example B. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
t = 1, v = 64, so
h = -16(1)2 + 64(1)
h = -16 + 64 = 48 ft
Hence after 1 second, the stone is 48 ft above the ground.

36. b. How long will it take for it to fall back to the ground?
2nd-Degree-Equation Word Problems

37. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
2nd-Degree-Equation Word Problems

38. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
2nd-Degree-Equation Word Problems

39. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
2nd-Degree-Equation Word Problems

40. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0
2nd-Degree-Equation Word Problems

41. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
2nd-Degree-Equation Word Problems

42. b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.
2nd-Degree-Equation Word Problems

43. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

44. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

45. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

46. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

47. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
= - 64 + 128
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

48. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
= - 64 + 128
= 64 (ft)
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

49. 2nd-Degree-Equation Word Problems
c. What is the maximum height obtained?
Since it takes 4 seconds for the stone to fall back to the
ground, at 2 seconds it must reach the highest point.
Hence t = 2, v = 64, need to find h.
h = -16(2)2 + 64(2)
= - 64 + 128
= 64 (ft)
Therefore, the stone reached the maximum height of 64 feet.
b. How long will it take for it to fall back to the ground?
To fall back to the ground means the height h is 0.
Hence
h = 0, v = 64, need to find t.
0 = -16 t2 + 64t
0 = -16t(t – 4)
t = 0 or t – 4 = 0 or t = 4
Therefore it takes 4 seconds.

50. 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree
equations.

51. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Formulas of area in mathematics also lead to 2nd degree
equations.

52. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
Formulas of area in mathematics also lead to 2nd degree
equations.

53. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
Formulas of area in mathematics also lead to 2nd degree
equations.

54. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.

55. If L and W are in a given unit, then A is in unit2.
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.

56. If L and W are in a given unit, then A is in unit2.
For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.

57. If L and W are in a given unit, then A is in unit2.
For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
1 in
1 in
1 in2
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.

58. If L and W are in a given unit, then A is in unit2.
For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2
1 in
1 in
1 in2
2 in
3 in
6 in2
2nd-Degree-Equation Word Problems
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
the area A of the rectangle is
A = LW.
L
w A = LW
Formulas of area in mathematics also lead to 2nd degree
equations.

59. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
2nd-Degree-Equation Word Problems
L
w A = LW

60. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
2nd-Degree-Equation Word Problems
L
w A = LW

61. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
2nd-Degree-Equation Word Problems
L
w A = LW

62. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
2nd-Degree-Equation Word Problems
L
w A = LW

63. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
2nd-Degree-Equation Word Problems
L
w A = LW

64. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
2nd-Degree-Equation Word Problems
L
w A = LW

65. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
2nd-Degree-Equation Word Problems
L
w A = LW

66. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
2nd-Degree-Equation Word Problems
L
w A = LW

67. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
L
w A = LW

68. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
L
w A = LW
Area of a parallelogram A = BH

69. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines.
L
w A = LW
Area of a parallelogram A = BH

70. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines.
If we move the shaded part as shown,
H=height
B=base
L
w A = LW
Area of a parallelogram A = BH

71. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines.
If we move the shaded part as shown,
H=height
B=base
L
w A = LW
Area of a parallelogram A = BH

72. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines.
If we move the shaded part as shown,
we reshaped the parallelogram
into a rectangle. B=base
H=height
L
w A = LW
Area of a parallelogram A = BH

73. Example C. The length of a rectangle is 4 inches more than
the width. The area is 21 in2. Find the length and width.
Let x = width, then the length = (x + 4)
LW = A, so (x + 4)x = 21
x2 + 4x = 21
x2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = - 7 or x = 3
Therefore, the width is 3 and the length is 7.
2nd-Degree-Equation Word Problems
Area of a Parallelogram
A parallelogram is the area enclosed
by two sets of parallel lines.
If we move the shaded part as shown,
we reshaped the parallelogram
into a rectangle. So the area A
of the parallelogram is A = BH
H=height
B=base
L
w A = LW
Area of a parallelogram A = BH

74. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
2nd-Degree-Equation Word Problems
2x + 3
x

75. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27

76. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0

77. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0

78. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft

79. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft

80. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Area of a triangle A = ½ BH

81. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
B
H
Area of a triangle A = ½ BH

82. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown .
Area of a triangle A = ½ BH

83. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
Area of a triangle A = ½ BH

84. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
We obtain a parallelogram.
Area of a triangle A = ½ BH

85. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
We obtain a parallelogram.
If A is the area of the triangle,
Area of a triangle A = ½ BH

86. Example D. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
2nd-Degree-Equation Word Problems
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
B
H
2x + 3
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
(2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft
Take another copy and place it
above the original one as shown.
We obtain a parallelogram.
If A is the area of the triangle,
then 2A = HB or .A = BH
2 Area of a triangle A = ½ BH

87. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
2nd-Degree-Equation Word Problems

88. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
2nd-Degree-Equation Word Problems

89. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
2nd-Degree-Equation Word Problems

90. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
2nd-Degree-Equation Word Problems
2x– 3
x

91. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
2nd-Degree-Equation Word Problems
2x– 3
x

92. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
2nd-Degree-Equation Word Problems
2x– 3
x

93. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
2nd-Degree-Equation Word Problems
2x– 3
x

94. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
2nd-Degree-Equation Word Problems
2x– 3
x

95. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
2nd-Degree-Equation Word Problems
2x– 3
x

96. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
2nd-Degree-Equation Word Problems
2x– 3
x

97. Example E. The base of a triangle is 3 inches shorter than
the twice of the height. The area is 10 in2. Find the base
and height.
Let x = height, then the base = (2x – 3)
Hence, use the formula 2A = BH
2*10 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
x = 4 or x = -5/2
Therefore the height is 4 in. and the base is 5 in.
2nd-Degree-Equation Word Problems
2x– 3
x

98. 2nd-Degree-Equation Word Problems
Coin Flips and Probability

99. 2nd-Degree-Equation Word Problems
Coin Flips and Probability
We have a coin and we like to determine the “chances” that
it would land on the head (H) or that tail (T) when it’s flipped.

100. 2nd-Degree-Equation Word Problems
Coin Flips and Probability
We flip the coin 100 times and 50 times it landed H’s so we
estimate the probability to get a H, or prob(H) = 50/100 = 0.5.
We have a coin and we like to determine the “chances” that
it would land on the head (H) or that tail (T) when it’s flipped.

101. 2nd-Degree-Equation Word Problems
Coin Flips and Probability
We flip the coin 100 times and 50 times it landed H’s so we
estimate the probability to get a H, or prob(H) = 50/100 = 0.5.
Note that it must have landed T’s 50 times so prob(T) = 0.5.
We have a coin and we like to determine the “chances” that
it would land on the head (H) or that tail (T) when it’s flipped.

102. 2nd-Degree-Equation Word Problems
Coin Flips and Probability
We flip the coin 100 times and 50 times it landed H’s so we
estimate the probability to get a H, or prob(H) = 50/100 = 0.5.
Note that it must have landed T’s 50 times so prob(T) = 0.5.
We say this coin is fair because we’ve a 50–50 chances
to obtain a head or tail on a flip.
We have a coin and we like to determine the “chances” that
it would land on the head (H) or that tail (T) when it’s flipped.

103. 2nd-Degree-Equation Word Problems
Coin Flips and Probability
We flip the coin 100 times and 50 times it landed H’s so we
estimate the probability to get a H, or prob(H) = 50/100 = 0.5.
Note that it must have landed T’s 50 times so prob(T) = 0.5.
We say this coin is fair because we’ve a 50–50 chances
to obtain a head or tail on a flip.
However if 60 of the 100 flips are H’s, so 40 flips are T’s,
then prob(H) = 60/100 = 0.6 and the prob(T) = 40/100 = 0.4.
We have a coin and we like to determine the “chances” that
it would land on the head (H) or that tail (T) when it’s flipped.

104. 2nd-Degree-Equation Word Problems
Coin Flips and Probability
We flip the coin 100 times and 50 times it landed H’s so we
estimate the probability to get a H, or prob(H) = 50/100 = 0.5.
Note that it must have landed T’s 50 times so prob(T) = 0.5.
We say this coin is fair because we’ve a 50–50 chances
to obtain a head or tail on a flip.
However if 60 of the 100 flips are H’s, so 40 flips are T’s,
then prob(H) = 60/100 = 0.6 and the prob(T) = 40/100 = 0.4.
We say this coin is unfair because it’s not a 50–50 split.
We have a coin and we like to determine the “chances” that
it would land on the head (H) or that tail (T) when it’s flipped.

105. 2nd-Degree-Equation Word Problems
Coin Flips and Probability
We flip the coin 100 times and 50 times it landed H’s so we
estimate the probability to get a H, or prob(H) = 50/100 = 0.5.
Note that it must have landed T’s 50 times so prob(T) = 0.5.
We say this coin is fair because we’ve a 50–50 chances
to obtain a head or tail on a flip.
However if 60 of the 100 flips are H’s, so 40 flips are T’s,
then prob(H) = 60/100 = 0.6 and the prob(T) = 40/100 = 0.4.
We say this coin is unfair because it’s not a 50–50 split.
Since H and T are the only possible outcomes for coin flips,
we must have that prob(H) + prob(T) = 1.
We have a coin and we like to determine the “chances” that
it would land on the head (H) or that tail (T) when it’s flipped.

106. 2nd-Degree-Equation Word Problems
Coin Flips and Probability
We flip the coin 100 times and 50 times it landed H’s so we
estimate the probability to get a H, or prob(H) = 50/100 = 0.5.
Note that it must have landed T’s 50 times so prob(T) = 0.5.
We say this coin is fair because we’ve a 50–50 chances
to obtain a head or tail on a flip.
However if 60 of the 100 flips are H’s, so 40 flips are T’s,
then prob(H) = 60/100 = 0.6 and the prob(T) = 40/100 = 0.4.
We say this coin is unfair because it’s not a 50–50 split.
Since H and T are the only possible outcomes for coin flips,
we must have that prob(H) + prob(T) = 1.
If prob(H) = p and prob(T) = q, then p + q = 1, so q = (1 – p).
(Hence if we know one, we know the other.)
We have a coin and we like to determine the “chances” that
it would land on the head (H) or that tail (T) when it’s flipped.

107. 2nd-Degree-Equation Word Problems
Double Flips

108. We flip a coin twice and note the sides it landed on in order.
2nd-Degree-Equation Word Problems
Double Flips

109. We flip a coin twice and note the sides it landed on in order.
There are four possible outcomes: HH, HT, TH and TT.
2nd-Degree-Equation Word Problems
Double Flips

110. We flip a coin twice and note the sides it landed on in order.
There are four possible outcomes: HH, HT, TH and TT.
2nd-Degree-Equation Word Problems
Double Flips
Fact: The probability of a double–flip outcomes is
the product of probabilities of each outcome*.

111. We flip a coin twice and note the sides it landed on in order.
There are four possible outcomes: HH, HT, TH and TT.
2nd-Degree-Equation Word Problems
Double Flips
Fact: The probability of a double–flip outcomes is
the product of probabilities of each outcome*.
For example if prob(H) = 0.6 and prob(T) = 0.4,

112. We flip a coin twice and note the sides it landed on in order.
There are four possible outcomes: HH, HT, TH and TT.
2nd-Degree-Equation Word Problems
Double Flips
Fact: The probability of a double–flip outcomes is
the product of probabilities of each outcome*.
For example if prob(H) = 0.6 and prob(T) = 0.4,
then prob(HH) = (0.6)(0.6) = 0.36,

113. We flip a coin twice and note the sides it landed on in order.
There are four possible outcomes: HH, HT, TH and TT.
2nd-Degree-Equation Word Problems
Double Flips
Fact: The probability of a double–flip outcomes is
the product of probabilities of each outcome*.
For example if prob(H) = 0.6 and prob(T) = 0.4,
then prob(HH) = (0.6)(0.6) = 0.36, prob(TT) = (0.4) (0.4) = 0.16,

114. We flip a coin twice and note the sides it landed on in order.
There are four possible outcomes: HH, HT, TH and TT.
2nd-Degree-Equation Word Problems
Double Flips
Fact: The probability of a double–flip outcomes is
the product of probabilities of each outcome*.
For example if prob(H) = 0.6 and prob(T) = 0.4,
then prob(HH) = (0.6)(0.6) = 0.36, prob(TT) = (0.4) (0.4) = 0.16,
and prob(HT) = prob(HT) = (0.6)(0.4) = 0.24

115. We flip a coin twice and note the sides it landed on in order.
There are four possible outcomes: HH, HT, TH and TT.
2nd-Degree-Equation Word Problems
Double Flips
Fact: The probability of a double–flip outcomes is
the product of probabilities of each outcome*.
For example if prob(H) = 0.6 and prob(T) = 0.4,
then prob(HH) = (0.6)(0.6) = 0.36, prob(TT) = (0.4) (0.4) = 0.16,
and prob(HT) = prob(HT) = (0.6)(0.4) = 0.24
* We are assuming here that the 1st flip will not influence,
the outcome of the 2nd flip (e.g. the 1st flip damaged the coin
and altered the probabilities of future flips.)

116. We flip a coin twice and note the sides it landed on in order.
There are four possible outcomes: HH, HT, TH and TT.
2nd-Degree-Equation Word Problems
Double Flips
Fact: The probability of a double–flip outcomes is
the product of probabilities of each outcome*.
For example if prob(H) = 0.6 and prob(T) = 0.4,
then prob(HH) = (0.6)(0.6) = 0.36, prob(TT) = (0.4) (0.4) = 0.16,
and prob(HT) = prob(HT) = (0.6)(0.4) = 0.24
* We are assuming here that the 1st flip will not influence,
the outcome of the 2nd flip (e.g. the 1st flip damaged the coin
and altered the probabilities of future flips.)
If the result of an event A (flipping a coin) would not influence
the result of another event B (flipping the coin 2nd time),
we say the events A and B are independent.
Here we are assuming the 1st flip is independent of the 2nd flip.

117. 2nd-Degree-Equation Word Problems
In general if prob(H) = p and prob(T) = q = (1 – p),
then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2
and that prob(HT) = prob(HT) = p * q = p(1 – p)
Double Flips

118. 2nd-Degree-Equation Word Problems
In general if prob(H) = p and prob(T) = q = (1 – p),
then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2
and that prob(HT) = prob(HT) = p * q = p(1 – p)
Let’s use these facts to find the probability assignment of a coin.
Double Flips

119. 2nd-Degree-Equation Word Problems
In general if prob(H) = p and prob(T) = q = (1 – p),
then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2
and that prob(HT) = prob(HT) = p * q = p(1 – p)
Let’s use these facts to find the probability assignment of a coin.
Example F. We have a coin and we want to estimate
the probability of getting a head when it’s flipped.
Out of 100 double–flips, exactly 16 double flips are HT.
Use this information to estimate p = prob(H).
Double Flips

120. 2nd-Degree-Equation Word Problems
In general if prob(H) = p and prob(T) = q = (1 – p),
then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2
and that prob(HT) = prob(HT) = p * q = p(1 – p)
Let’s use these facts to find the probability assignment of a coin.
Example F. We have a coin and we want to estimate
the probability of getting a head when it’s flipped.
Out of 100 double–flips, exactly 16 double flips are HT.
Use this information to estimate p = prob(H).
Out of 100 double–flips, exactly 16 double flips are HT
means that prob(HT) = 16/100 or 4/25.
Double Flips

121. 2nd-Degree-Equation Word Problems
In general if prob(H) = p and prob(T) = q = (1 – p),
then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2
and that prob(HT) = prob(HT) = p * q = p(1 – p)
Let’s use these facts to find the probability assignment of a coin.
Example F. We have a coin and we want to estimate
the probability of getting a head when it’s flipped.
Out of 100 double–flips, exactly 16 double flips are HT.
Use this information to estimate p = prob(H).
Out of 100 double–flips, exactly 16 double flips are HT
means that prob(HT) = 16/100 or 4/25.
Since prob(HT) = pq = p(1 – p) = 4/25 so that p – p2 = 4/25.
Double Flips

122. 2nd-Degree-Equation Word Problems
In general if prob(H) = p and prob(T) = q = (1 – p),
then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2
and that prob(HT) = prob(HT) = p * q = p(1 – p)
Let’s use these facts to find the probability assignment of a coin.
Example F. We have a coin and we want to estimate
the probability of getting a head when it’s flipped.
Out of 100 double–flips, exactly 16 double flips are HT.
Use this information to estimate p = prob(H).
Out of 100 double–flips, exactly 16 double flips are HT
means that prob(HT) = 16/100 or 4/25.
Since prob(HT) = pq = p(1 – p) = 4/25 so that p – p2 = 4/25.
Multiple both sides by 25, we get 25p – 25p2 = 4.
Double Flips

123. 2nd-Degree-Equation Word Problems
In general if prob(H) = p and prob(T) = q = (1 – p),
then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2
and that prob(HT) = prob(HT) = p * q = p(1 – p)
Let’s use these facts to find the probability assignment of a coin.
Example F. We have a coin and we want to estimate
the probability of getting a head when it’s flipped.
Out of 100 double–flips, exactly 16 double flips are HT.
Use this information to estimate p = prob(H).
Out of 100 double–flips, exactly 16 double flips are HT
means that prob(HT) = 16/100 or 4/25.
Since prob(HT) = pq = p(1 – p) = 4/25 so that p – p2 = 4/25.
Multiple both sides by 25, we get 25p – 25p2 = 4.
Hence 0 = 25p2 – 25p + 4 or 0 = (5p – 4)(5p – 1),
Double Flips

124. 2nd-Degree-Equation Word Problems
In general if prob(H) = p and prob(T) = q = (1 – p),
then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2
and that prob(HT) = prob(HT) = p * q = p(1 – p)
Let’s use these facts to find the probability assignment of a coin.
Example F. We have a coin and we want to estimate
the probability of getting a head when it’s flipped.
Out of 100 double–flips, exactly 16 double flips are HT.
Use this information to estimate p = prob(H).
Out of 100 double–flips, exactly 16 double flips are HT
means that prob(HT) = 16/100 or 4/25.
Since prob(HT) = pq = p(1 – p) = 4/25 so that p – p2 = 4/25.
Multiple both sides by 25, we get 25p – 25p2 = 4.
Hence 0 = 25p2 – 25p + 4 or 0 = (5p – 4)(5p – 1),
therefore prob(H) = p = 4/5 or p = 1/5.
Double Flips

125. Exercise A. Use the formula h = –16t2 + vt for the following
problems.
2nd-Degree-Equation Word Problems
1. A stone is thrown upward at a speed of v = 64 ft/sec,
how long does it take for it’s height to reach 48 ft?
2. A stone is thrown upward at a speed of v = 64 ft/sec,
how long does it take for it’s height to reach 28 ft?
3. A stone is thrown upward at a speed of v = 96 ft/sec,
a. how long does it take for its height to reach 80 ft? Draw a
picture.
b. how long does it take for its height to reach the highest
point?
c. What is the maximum height it reached?
4. A stone is thrown upward at a speed of v = 128 ft/sec,
a. how long does it take for its height to reach 256 ft?
Draw a picture. How long does it take for its height to reach
the highest point and what is the maximum height it reached?

126. B. Given the following area measurements, find x.
2nd-Degree-Equation Word Problems
5.
8 ft2
x + 2
x
6.
12 ft2
x
(x – 1)
7.
x + 2
8.
12 ft2 x
(x + 4)
9.
24 ft2
(3x – 1)
x
10.
15 ft2x
18 ft2 x
(4x + 1)

127. B. Given the following area measurements, find x.
2nd-Degree-Equation Word Problems
2x + 1
11.
x5cm2
12.
x
2x – 3
9cm2
2x + 1
13.
x
18km2
14.
(x + 3)
(5x + 3)
24km2
15. 16.
16km2
2
x
x + 1
35km2
2
x
2x – 1