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# 5 82nd-degree-equation word problems

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## 5 82nd-degree-equation word problemsPresentation Transcript

• 2nd-Degree-Equation Word Problems
• 2nd-Degree-Equation Word Problems
Many physics formulas are 2nd degree.
• 2nd-Degree-Equation Word Problems
Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
• 2nd-Degree-Equation Word Problems
Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
where
h = height in feet
t = time in second
v = upward speed in feet per second
• 2nd-Degree-Equation Word Problems
Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
height = -16t2 + vt
after t seconds
where
h = height in feet
t = time in second
v = upward speed in feet per second
• 2nd-Degree-Equation Word Problems
Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
height = -16t2 + vt
after t seconds
where
h = height in feet
t = time in second
v = upward speed in feet per second
Example A. If a stone is thrown straight
up at a speed of 64 ft per second,
a. how high is it after 1 second?
• 2nd-Degree-Equation Word Problems
Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
height = -16t2 + vt
after t seconds
where
h = height in feet
t = time in second
v = upward speed in feet per second
Example A. 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,
• 2nd-Degree-Equation Word Problems
Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
height = -16t2 + vt
after t seconds
where
h = height in feet
t = time in second
v = upward speed in feet per second
Example A. 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)
• 2nd-Degree-Equation Word Problems
Many physics formulas are 2nd degree.
If a stone is thrown straight up on Earth then
h = -16t2 + vt
height = -16t2 + vt
after t seconds
where
h = height in feet
t = time in second
v = upward speed in feet per second
Example A. 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
• 2nd-Degree-Equation Word Problems
b. How long will it take for it to fall back to the ground?
• 2nd-Degree-Equation Word Problems
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
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
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
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
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
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
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
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.
c. What is the maximum height obtained?
• 2nd-Degree-Equation Word Problems
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.
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.
• 2nd-Degree-Equation Word Problems
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.
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.
• 2nd-Degree-Equation Word Problems
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.
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)
• 2nd-Degree-Equation Word Problems
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.
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
• 2nd-Degree-Equation Word Problems
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.
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)
• 2nd-Degree-Equation Word Problems
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.
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.
• 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree equations.
• 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree equations.
Area of a Rectangle
• 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree equations.
Area of a Rectangle
Given a rectangle, let
L = length of a rectangle
W = width of the rectangle,
• 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree equations.
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.
• 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree equations.
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
A = LW
w
• 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree equations.
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
A = LW
w
If L and W are in a given unit, then A is in unit2.
• 2nd-Degree-Equation Word Problems
Formulas of area in mathematics also lead to 2nd degree equations.
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
A = LW
w
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
Formulas of area in mathematics also lead to 2nd degree equations.
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
A = LW
w
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
Formulas of area in mathematics also lead to 2nd degree equations.
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
A = LW
w
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
3 in
1 in
1 in
2 in
1 in2
6 in2
• 2nd-Degree-Equation Word Problems
Example B. 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
Example B. 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
Example B. 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
Example B. 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
Example B. 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
Example B. 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
Example B. 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
Example B. 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
Example B. 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
Example B. 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.
Area of a Parallelogram
H=height
B=base
• 2nd-Degree-Equation Word Problems
Example B. 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.
Area of a Parallelogram
A parallelogram is the area enclosed by two sets of parallel lines.
H=height
B=base
• 2nd-Degree-Equation Word Problems
Example B. 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.
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
• 2nd-Degree-Equation Word Problems
Example B. 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.
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
• 2nd-Degree-Equation Word Problems
Example B. 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.
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.
H=height
B=base
• 2nd-Degree-Equation Word Problems
Example B. 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.
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
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
x
2x + 3
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x + 3
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0
2x + 3
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
Area of a Triangle
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
H
B
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
Take another copy and place it above the original one as shown .
H
B
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
Take another copy and place it above the original one as shown.
H
B
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
Take another copy and place it above the original one as shown.
We obtain a parallelogram.
H
B
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
Take another copy and place it above the original one as shown.
We obtain a parallelogram.
H
If A is the area of the triangle,
B
• 2nd-Degree-Equation Word Problems
Example C. The area of the parallelogram shown is 27 ft2.
Find x.
The area is (base)(height) or that
x(2x + 3) = 27
x
2x2 + 3x = 27
2x2 + 3x – 27 = 0 Factor
2x + 3
(2x + 9)(x – 3) = 0
x = –9/2, x = 3 ft
Area of a Triangle
Given the base (B) and the height (H) of a triangle as shown.
Take another copy and place it above the original one as shown.
We obtain a parallelogram.
H
If A is the area of the triangle,
then 2A = HB or .
BH
B
A =
2
• 2nd-Degree-Equation Word Problems
Example D. 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
Example D. 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
Example D. 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 2nd version of the formula we have
20 = (2x – 3) x
• 2nd-Degree-Equation Word Problems
Example D. 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 2nd version of the formula we have
20 = (2x – 3) x
20 = 2x2 – 3x
• 2nd-Degree-Equation Word Problems
Example D. 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 2nd version of the formula we have
20 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
• 2nd-Degree-Equation Word Problems
Example D. 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 2nd version of the formula we have
20 = (2x – 3) x
20 = 2x2 – 3x
0 = 2x2 – 3x – 20
0 = (x – 4)(2x + 5)
• 2nd-Degree-Equation Word Problems
Example D. 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 2nd version of the formula we have
20 = (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
Example D. 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
Example D. 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
Example D. 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
Example D. 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)
x
2x– 3
• 2nd-Degree-Equation Word Problems
Example D. 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
x
2x– 3
• 2nd-Degree-Equation Word Problems
Example D. 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
x
2x– 3
• 2nd-Degree-Equation Word Problems
Example D. 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
x
2x– 3
• 2nd-Degree-Equation Word Problems
Example D. 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
2x– 3
• 2nd-Degree-Equation Word Problems
Example D. 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
x
2x– 3
• 2nd-Degree-Equation Word Problems
Example D. 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
x
2x– 3
• 2nd-Degree-Equation Word Problems
Example D. 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.
x
2x– 3