14 2nd degree-equation word problems

2nd-Degree-Equation Word Problems

Many physics formulas are 2nd degree.


Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then

h = -16t2 + vt



h = -16t2 + vt


whereh = height in feett = time in secondv = upward speed in feet per second


h = -16t2 + vt


height = -16t2 + vt after t secondswhere

h = height in feett = time in secondv = upward speed in feet per second


h = -16t2 + vt




Example A. If a stone is thrown straightup at a speed of 64 ft per second,

a. how high is it after 1 second?


h = -16t2 + vt





a. how high is it after 1 second? t = 1, v = 64,


h = -16t2 + vt





a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1)


h = -16t2 + vt





a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1) h = -16 + 64 = 48 ft

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.


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.


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


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)


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


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


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?



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.



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.



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)



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



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)



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 ProblemsFormulas of area in mathematics also lead to 2nd degree equations.


Area of a Rectangle

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,



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.



Area of a Rectangle


L

w A = LW


If L and W are in a given unit, then A is in unit2.


Area of a Rectangle


L

w A = LW


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


Area of a Rectangle


L

w A = LW



1 in

1 in

1 in2


Area of a Rectangle


L

w A = LW



1 in

1 in

1 in2

2 in

3 in

6 in2


Area of a Rectangle


L

w A = LW


Example B. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.


L

w A = LW

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)


L

w A = LW

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


L

w A = LW

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


L

w A = LW

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


L

w A = LW

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


L

w A = LW

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


L

w A = LW

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 = 3Therefore, the width is 3 and the length is 7.


L

w A = LW



Area of a Parallelogram

B=base

H=height

L

w A = LW




A parallelogram is the area enclosed by two sets of parallel lines.

B=base

H=height

L

w A = LW




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




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




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 Aof the parallelogram is A = BH.

H=height

B=base

L

w A = LW

Example C. The area of the parallelogram shown is 27 ft2.Find x.


2x + 3

x

Example C. The area of the parallelogram shown is 27 ft2.Find x.The area is (base)(height) or thatx(2x + 3) = 27


2x + 3

x2x2 + 3x = 27



2x + 3

x2x2 + 3x = 27 2x2 + 3x – 27 = 0



2x + 3

x2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0



2x + 3

x2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft



Area of a Triangle

2x + 3




Area of a Triangle

Given the base (B) and the height (H) of a triangle as shown.

2x + 3


B

H



Area of a Triangle


B

H

2x + 3


Take another copy and place it above the original one as shown .



Area of a Triangle


B

H

2x + 3


Take another copy and place it above the original one as shown.



Area of a Triangle


B

H

2x + 3


Take another copy and place it above the original one as shown.We obtain a parallelogram.



Area of a Triangle


B

H

2x + 3


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


B

H

2x + 3


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

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.


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)


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)


2x– 3

x

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


2x– 3

x

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) x20 = 2x2 – 3x


2x– 3

x

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) x20 = 2x2 – 3x0 = 2x2 – 3x – 20


2x– 3

x

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) x20 = 2x2 – 3x0 = 2x2 – 3x – 20 0 = (x – 4)(2x + 5)


2x– 3

x

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) x20 = 2x2 – 3x0 = 2x2 – 3x – 20 0 = (x – 4)(2x + 5)x = 4 or x = -5/2


2x– 3

x

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) x20 = 2x2 – 3x0 = 2x2 – 3x – 20 0 = (x – 4)(2x + 5)x = 4 or x = -5/2Therefore the height is 4 in. and the base is 5 in.


2x– 3

x

Exercise A. Use the formula h = –16t2 + vt for the following 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?

B. Given the following area measurements, find x.


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)

B. Given the following area measurements, find x.


2x + 1

11.

x5cm2

12.

x

2x – 3

9cm2

2x + 1

13.

x18km2

14.

(x + 3)

(5x + 3)24km2

15. 16.16km2

2

x

x + 1

35km22

x

2x – 1

14 2nd degree-equation word problems

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Transcript of 14 2nd degree-equation word problems