1. 2nd-Degree-Equation Word Problems

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

3. 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 related linearly,
then their product is a 2nd degree expression.
2nd-Degree-Equation Word Problems

4. 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 related linearly,
then their product is a 2nd degree expression.
2nd-Degree-Equation Word Problems

5. 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 related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
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.

6. 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 related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
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.

7. 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 related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
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).
Their product of them is 96 so
x(3x – 2) = 96

8. 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 related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
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).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96

9. 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 related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
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).
Their product of them is 96 so
x(3x – 2) = 96 expand
3x2 – 2x = 96
3x2 – 2x – 96 = 0

10. 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 related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
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).
Their 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.

11. 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 related linearly,
then their product is a 2nd degree expression.
Hence the problem of finding A and B when their product
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).
Their 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.

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

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

14. 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

15. 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

16. 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?

17. 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,

18. 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)

19. 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

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

21. 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

22. 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.
2nd-Degree-Equation Word Problems

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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.

29. 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.

30. 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.

31. 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.

32. 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.

33. 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.

34. 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 maximum height is 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.

35. 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree
equations.

36. 2nd-Degree-Equation Word Problems
Area of a Rectangle
Formulas of area in mathematics also lead to 2nd degree
equations.

37. 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.

38. 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.

39. 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.

40. 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.

41. 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.

42. 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.

43. 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.

44. 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

45. 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

46. 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

47. 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

48. 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

49. 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

50. 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

51. 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

52. 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

53. 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
B=base
H=height
L
w A = LW

54. 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.
B=base
H=height
L
w A = LW

55. 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

56. 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

57. 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
get a rectangle.
B=base
H=height
L
w A = LW

58. 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
get a rectangle. Hence the area A
of the parallelogram is A = BH.
H=height
B=base
L
w A = LW

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

60. 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

61. 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

62. 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

63. 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

64. 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

65. 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

66. 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

67. 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 .

68. 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.

69. 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.

70. 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,

71. 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

72. 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

73. 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

74. 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

75. 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

76. 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

77. 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

78. 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

79. 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

80. 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

81. 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

82. 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

83. 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?

84. 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)

85. 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