Chapter 13 355 CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS Chapter Objectives By the end of this chapter, students should be able to: Apply the Square Root Property to solve quadratic equations Solve quadratic equations by completing the square and using the Quadratic Formula Solve applications by applying the quadratic formula or completing the square Contents CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS .................................................................. 355 SECTION 13.1: THE SQUARE ROOT PROPERTY .................................................................................... 356 A. SOLVE BASIC QUADRATIC EQUATIONS USING SQUAREROOT PROPERTY ............................. 356 B. ISOLATE THE SQUARED TERM.................................................................................................. 358 C. USE THE PERFECT SQUARE FORMULA ..................................................................................... 359 EXERCISE ........................................................................................................................................... 360 SECTION 13.2: COMPLETING THE SQUARE .......................................................................................... 361 A. COMPLETE THE SQUARE .......................................................................................................... 361 B. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, a = 1 .................................. 362 C. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, A ≠ 1.................................. 363 EXERCISE ........................................................................................................................................... 365 SECTION 13.3: QUADRATIC FORMULA ................................................................................................ 366 A. DETERMINANT OF A QUADRATIC EQUATION ......................................................................... 366 B. APPLY THE QUADRATIC FORMULA .......................................................................................... 368 C. MAKE ONE SIDE OF AN EQUATION EQUAL TO ZERO .............................................................. 370 EXERCISE ........................................................................................................................................... 371 SECTION 13.4: APPLICATIONS WITH QUADRATIC EQUATIONS .......................................................... 372 A. PYTHAGOREAN THEOREM ....................................................................................................... 372 B. PROJECTILE MOTION ................................................................................................................ 373 C. COST AND REVENUE ................................................................................................................. 374 EXERCISE ........................................................................................................................................... 376 CHAPTER REVIEW ................................................................................................................................. 377
Transcript
Chapter 13
355
CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS Chapter Objectives
By the end of this chapter, students should be able to: Apply the Square Root Property to solve quadratic equations Solve quadratic equations by completing the square and using the Quadratic Formula Solve applications by applying the quadratic formula or completing the square
Contents CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS .................................................................. 355
SECTION 13.1: THE SQUARE ROOT PROPERTY .................................................................................... 356
A. SOLVE BASIC QUADRATIC EQUATIONS USING SQUAREROOT PROPERTY ............................. 356
B. ISOLATE THE SQUARED TERM .................................................................................................. 358
C. USE THE PERFECT SQUARE FORMULA ..................................................................................... 359
We might recognize a quadratic equation from the factoring chapter as a trinomial equation. Although, it may seem that they are the same, but they aren’t the same. Trinomial equations are equations with any three terms. These terms can be any three terms where the degree of each can vary. On the other hand, quadratic equations are equations with specific degree on each term.
Definition
A quadratic equation is a polynomial equation of the form 𝒂𝒂𝒂𝒂𝟐𝟐 + 𝒃𝒃𝒂𝒂 + 𝒄𝒄 = 𝟎𝟎
Where 𝒂𝒂𝒂𝒂𝟐𝟐 is called the leading term, 𝒃𝒃𝒂𝒂 is call the linear term, and 𝒄𝒄 is called the constant coefficient (or constant term). Additionally, 𝒂𝒂 ≠ 𝟎𝟎.
SECTION 13.1: THE SQUARE ROOT PROPERTY A. SOLVE BASIC QUADRATIC EQUATIONS USING SQUAREROOT PROPERTY
Square root property
Let 𝒂𝒂 ≥ 𝟎𝟎 and 𝒂𝒂 ≥ 𝟎𝟎. Then
𝒂𝒂𝟐𝟐 = 𝒂𝒂 if and only if 𝒂𝒂 = ±√𝒂𝒂 In other words,
𝒂𝒂𝟐𝟐 = 𝒂𝒂 if and only if 𝒂𝒂 = √𝒂𝒂 or 𝒂𝒂 = −√𝒂𝒂
MEDIA LESSON Solve basic quadratic equations using square root property (Duration 2:53)
View the video lesson, take notes and complete the problems below
B. ISOLATE THE SQUARED TERM Let’s look at examples where the leading term, or squared term, is not isolated. Recall, the squared term must be isolated to apply the square root property.
MEDIA LESSON Solve equations using square root property – Isolating the squared term 1st (Duration 5:00)
View the video lesson, take notes and complete the problems below
Before we can clear an exponent, it must first be _____________________________.
Example:
a) 4 − 2(2𝑥𝑥 + 1)2 = −46 b) 5(3𝑥𝑥 − 2)2 + 6 = 46
YOU TRY
Solve.
a) 5(3x − 6)2 + 7 = 27
b) 5(r + 4)2 + 1 = 626
Note: When we have the other side of the equation of a squared term negative, the equation does not have a real solution. For example, the equation 𝑥𝑥2 = −1 does not have a real solution. There is a complex solution for this equation but we will not discuss it in this class.
C. USE THE PERFECT SQUARE FORMULA In order for us to be able to apply the square root property to solve a quadratic equation, we cannot have the 𝑥𝑥 term in the middle because if we apply the square root property to the 𝑥𝑥 term, we will make the equation more complicated to solve. However sometimes, we have special cases that we can apply the perfect square formula to get rid of the 𝑥𝑥 term in the middle and then apply the square root property to solve the equations. Recall: Perfect square formula 𝒂𝒂𝟐𝟐 + 𝟐𝟐𝒂𝒂𝒃𝒃+ 𝒃𝒃𝟐𝟐 = (𝒂𝒂 + 𝒃𝒃)𝟐𝟐 or 𝒂𝒂𝟐𝟐 − 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐 = (𝒂𝒂 − 𝒃𝒃)𝟐𝟐
MEDIA LESSON Solve equations using square root property – Perfect Square formula (Duration 4:09)
View the video lesson, take notes and complete the problems below Example:
EXERCISE Solve by applying the square root property.
1) (𝑥𝑥 − 3)2 = 16
2) (𝑥𝑥 − 2)2 = 49
3) (𝑥𝑥 − 7)2 = 4
4) (𝑠𝑠 − 5)2 = 16
5) (𝑝𝑝 + 5)2 = 81
6) (𝑠𝑠 + 3)2 = 4
7) (𝑡𝑡 + 9)2 = 37
8) (𝑎𝑎 + 5)2 = 57
9) (𝑛𝑛 − 9)2 = 63
10) (𝑟𝑟 + 1)2 = 125
11) (9𝑟𝑟 + 1)2 = 9
12) (7𝑚𝑚 − 8)2 = 36
13) (3𝑠𝑠 − 6)2 = 25
14) 5(𝑘𝑘 − 7)2 − 6 = 369
15) 5(𝑔𝑔 − 5)2 + 13 = 103
16) 2𝑛𝑛2 + 7 = 5
17) (2𝑠𝑠 + 1)2 = 0
18) (𝑧𝑧 − 4)2 = 25
19) 3𝑛𝑛2 + 2𝑛𝑛 = 2𝑛𝑛 + 24
20) 8𝑛𝑛2 − 29 = 25 + 2𝑛𝑛2
21) 2(𝑟𝑟 + 9)2 − 19 = 37
22) 3(𝑛𝑛 − 3)2 + 2 = 164
23) 7(2𝑥𝑥 + 6)2 − 5 = 170
24) 6(4𝑥𝑥 − 4)2 − 5 = 145
Apply the perfect square formula and solve the equations by using the square root property.
25) 𝑥𝑥2 + 12𝑥𝑥 + 36 = 49
26) 𝑥𝑥2 + 6𝑥𝑥 + 9 = 2
27) 16𝑥𝑥2 − 40𝑥𝑥 + 25 = 16
28) 𝑥𝑥2 + 4𝑥𝑥 + 4 = 1
29) 𝑥𝑥2 − 14𝑥𝑥 + 49 = 9
30) 25𝑥𝑥2 + 10𝑥𝑥 + 1 = 49
Chapter 13
361
SECTION 13.2: COMPLETING THE SQUARE
When solving quadratic equations previously (then known as trinomial equations), we factored to solve. However, recall, not all equations are factorable. Consider the equation 𝑥𝑥2 − 2𝑥𝑥 − 7 = 0. This equation is not factorable, but there are two solutions to this equation: 1 + √2 and 1 − √2. Looking at the form of these solutions, we obtained these types of solutions in the previous section while using the square root property. If we can obtain a perfect square, then we can apply the square root property and solve as usual. This method we use to obtain a perfect square is called completing the square.
Recall. Special product formulas for perfect square trinomials:
We use these formulas to help us solve by completing the square.
A. COMPLETE THE SQUARE We first begin with completing the square and rewriting the trinomial in factored form using the perfect square trinomial formulas.
MEDIA LESSON Complete the square (Duration 5:00)
View the video lesson, take notes and complete the problems below
Complete the square. Find 𝒄𝒄.
𝒂𝒂𝟐𝟐 + 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐 is easily factored to ________________________________
To make 𝒂𝒂𝟐𝟐 + 𝒃𝒃𝒂𝒂+ 𝒄𝒄 a perfect square, 𝒄𝒄 = ___________________
To complete the square of any trinomial, we always square half of the linear term’s coefficient, i.e.,
�𝒃𝒃𝟐𝟐�𝟐𝟐
𝒐𝒐𝒐𝒐 �𝟏𝟏𝟐𝟐𝒃𝒃�
𝟐𝟐
We usually use the second expression when the middle term’s coefficient is a fraction.
YOU TRY
Complete the square by finding 𝒄𝒄:
a) 𝑥𝑥2 + 8𝑥𝑥 + 𝑐𝑐
b) 𝑥𝑥2 − 7𝑥𝑥 + 𝑐𝑐
c) 𝑥𝑥2 + 53𝑥𝑥 + 𝑐𝑐
B. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, a = 1
Steps to solving quadratic equations by completing the square
Given a quadratic equation 𝒂𝒂𝒂𝒂𝟐𝟐 + 𝒃𝒃𝒂𝒂 + 𝒄𝒄 = 𝟎𝟎, we can use the following method to solve for 𝒂𝒂.
Step 1. Rewrite the quadratic equation so that the coefficient of the leading term is one, and the original constant coefficient is on the opposite side of the equal sign from the leading and linear terms. 𝒂𝒂𝒂𝒂𝟐𝟐 + 𝒃𝒃𝒂𝒂 __________ = 𝒄𝒄 + _____________
Step 2. If 𝒂𝒂 ≠ 𝟏𝟏, divide both sides of the equation by 𝒂𝒂
Step 3. Complete the square, i.e., �𝒃𝒃𝟐𝟐�𝟐𝟐𝑜𝑜𝑟𝑟 �𝟏𝟏
𝟐𝟐𝒃𝒃�
𝟐𝟐 and add the result to both sides of the quadratic
equation. Step 4. Rewrite the perfect square trinomial in factored form. Step 5. Solve using the square root property. Step 6. Verify the solution(s).
MEDIA LESSON Solve quadratic equation by completing the square (Duration 8:40)
View the video lesson, take notes and complete the problems below
Solve the quadratic equation using the square root principle.
Solve each equation by completing the square. If the solution is not a real solution, then state not a real solution.
9) 6𝑟𝑟2 + 12𝑟𝑟 − 24 = −6
10) 6𝑛𝑛2 − 12𝑛𝑛 − 14 = 4
11) 𝑎𝑎2 − 56 = 10𝑎𝑎
12) 𝑥𝑥2 + 8𝑥𝑥 + 15 = 8
13) 8𝑛𝑛2 + 16𝑛𝑛 = 64
14) 𝑛𝑛2 + 4𝑛𝑛 = 12
15) 𝑥𝑥2 − 16𝑥𝑥 + 55 = 0
16) 𝑛𝑛2 = −21 + 10𝑛𝑛
17) 4𝑏𝑏2 − 15𝑏𝑏 + 56 = 3𝑏𝑏2
18) 𝑏𝑏2 + 7𝑏𝑏 − 33 = 0
19) 𝑥𝑥2 + 10𝑥𝑥 − 57 = 4
20) 𝑛𝑛2 − 8𝑛𝑛 − 12 = 0
21) 𝑛𝑛2 − 16𝑛𝑛 + 67 = 4
22) 𝑥𝑥2 = −10𝑥𝑥 − 29
23) 5𝑘𝑘2 − 10𝑘𝑘 + 48 = 0
24) 7𝑛𝑛2 − 𝑛𝑛 + 7 = 7𝑛𝑛 + 6𝑛𝑛2
25) 2𝑥𝑥2 + 4𝑥𝑥 + 38 = −6 26) 8𝑏𝑏2 + 16𝑏𝑏 − 37 = 5
27) 5𝑥𝑥2 + 5𝑥𝑥 = −31− 5𝑥𝑥 28) 𝑣𝑣2 + 5𝑣𝑣 + 28 = 0
29) 𝑘𝑘2 − 7𝑘𝑘 + 50 = 3
30) 5𝑥𝑥2 + 8𝑥𝑥 − 40 = 8
31) 8𝑟𝑟2 + 10𝑟𝑟 = −55
32) −2𝑥𝑥2 + 3𝑥𝑥 − 5 = −4𝑥𝑥2
33) 8𝑎𝑎2 + 16𝑎𝑎 − 1 = 0
34) 𝑝𝑝2 − 16𝑝𝑝 − 52 = 0
Chapter 13
366
SECTION 13.3: QUADRATIC FORMULA The quadratic formula is derived from the method of completing the square. If we took a general quadratic equation
𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0
and solved for 𝑥𝑥 by completing the square, we would obtain the quadratic formula. Let’s try this.
MEDIA LESSON Deriving the Quadratic Formula (Duration 4:04)
Quadratic formula
Given the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0. Then
𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑐𝑐
2𝑎𝑎
is called the quadratic formula. The quadratic formula is a formula for solving quadratic equations in terms of the coefficients.
A. DETERMINANT OF A QUADRATIC EQUATION To make the quadratic formula easier to manage, we should calculate the discriminant first. Then substitute it into the quadratic formula to find 𝑥𝑥.
Discriminant
Given the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, then its discriminant, denoted as the upper-case
Greek letter delta, Δ, is defined as 𝚫𝚫 = 𝒃𝒃𝟐𝟐 − 𝟒𝟒𝒂𝒂𝒄𝒄
√𝚫𝚫 = �𝒃𝒃𝟐𝟐 − 𝟒𝟒𝒂𝒂𝒄𝒄
Quadratic formula:
𝒂𝒂 =−𝒃𝒃 ± √𝚫𝚫
𝟐𝟐𝒂𝒂
Case 1: If Δ is positive, the equation has 2 solutions.
Case 2: If Δ is zero, the equation has 1 solution.
Case 3: If Δ is negative, the equation has no real solution. (The equation has two complex solutions
but we will not discuss complex numbers in this class).
EXERCISE Find the determinant and solve each equation by applying the quadratic formula. If the solution is not a real solution, then state not a real solution.
1) 2𝑥𝑥2 − 8𝑥𝑥 − 2 = 0 2) 3𝑟𝑟2 − 2𝑟𝑟 − 1 = 0
3) 2𝑥𝑥2 + 5𝑥𝑥 = −3
4) 𝑣𝑣2 − 4𝑣𝑣 − 5 = −8
5) 2𝑎𝑎2 + 3𝑎𝑎 + 14 = 6
6) 3𝑘𝑘2 + 3𝑘𝑘 − 4 = 7
7) 7𝑥𝑥2 + 3𝑥𝑥 − 16 = −2
8) 2𝑝𝑝2 + 6𝑝𝑝 − 16 = 4
9) 3𝑛𝑛2 + 3𝑛𝑛 = −3
10) 2𝑥𝑥2 = −7𝑥𝑥 + 49
11) 5𝑥𝑥2 = 7𝑥𝑥 + 7
12) 8𝑛𝑛2 = −3𝑛𝑛 − 8
13) 4𝑝𝑝2 + 5𝑝𝑝 − 36 = 3𝑝𝑝2
14) −5𝑛𝑛2 − 3𝑛𝑛 − 52 = 2 − 7𝑛𝑛2
15) 7𝑟𝑟2 − 12 = −3𝑟𝑟
16) 2𝑥𝑥2 + 4𝑥𝑥 + 12 = 8
17) 5𝑝𝑝2 + 2𝑝𝑝 + 6 = 0
18) 2𝑥𝑥2 − 2𝑥𝑥 − 15 = 0
19) 6𝑛𝑛2 − 3𝑛𝑛 + 3 = −4
20) 𝑚𝑚2 + 4𝑚𝑚 − 48 = −3
21) 4𝑛𝑛2 + 5𝑛𝑛 = 7
22) 3𝑟𝑟2 + 4 = −6𝑟𝑟
23) 3𝑏𝑏2 − 3 = 8𝑏𝑏
24) 6𝑣𝑣2 = 4 + 6𝑣𝑣
25) 6𝑎𝑎2 = −5𝑎𝑎 + 13
26) 2𝑘𝑘2 + 6𝑘𝑘 − 16 = 2𝑘𝑘
27) 6𝑏𝑏2 = 𝑏𝑏2 + 7 − 𝑏𝑏
28) 49𝑚𝑚2 − 28𝑚𝑚 + 4 = 0
29) 12𝑥𝑥2 + 𝑥𝑥 + 7 = 5𝑥𝑥2 + 5𝑥𝑥
30) 𝑥𝑥2 − 6𝑥𝑥 + 9 = 0
Chapter 13
372
SECTION 13.4: APPLICATIONS WITH QUADRATIC EQUATIONS There are many applications involving quadratic equations that it is almost challenging to pick the few equations to discuss for this section. Yet, we only choose a few applications that are common in most algebra classes. We start with the very famous Pythagorean Theorem.
A. PYTHAGOREAN THEOREM A long time ago, a Greek mathematician named Pythagoras discovered an interesting property about right triangles: the sum of the squares of the lengths of each of the triangle’s legs is the same as the square of the length of the triangle’s hypotenuse. This property- which has many applications in science, art, engineering, and architecture – is now called the Pythagorean Theorem.
Pythagorean Theorem
If 𝒂𝒂 and 𝒃𝒃 are the lengths of the legs of a right triangle and 𝒄𝒄 is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse:
a) Find the length of the hypotenuse of a right triangle given lengths 6 cm and 13 cm for the legs. Round to two decimal places.
b) Find the length of the missing leg of a right triangle given the length of the hypotenuse is 15 km and the length of the other leg is 11 km. Round to two decimal places.
.
B. PROJECTILE MOTION
MEDIA LESSON Quadratic equation application – Projectile motion (Duration 9:26)
View the video lesson, take notes and complete the problems below
Example:
You launch a toy rocket from a height of 5 feet. The height (h, in feet) of the rocket 𝑡𝑡 seconds after taking off is given by the formula ℎ = −3𝑡𝑡2 + 14𝑡𝑡 + 5.
a) How long will it take for the rocket to hit the ground? b) Find the time when the rocket is 5 feet from hitting the ground.
a) A rocket is launched at 𝑡𝑡 = 0 seconds. Its height, in meters above sea-level, is given by the equation ℎ = −4.9𝑡𝑡2 + 52𝑡𝑡 + 367
At what time does the rocket hit the ground? (Round answer to 2 decimal places.) Note: With applications, such as these, it doesn’t make sense to leave an exact answer, but better to
approximate. It is common language to say the rocket would hit the ground in decimal form.
C. COST AND REVENUE
Cost and Revenue
The cost refers to the amount of money the manufacturer should pay to produce a commodity. The revenue refers to the amount of money the consumer (or customer) pays for a commodity. Note, profit and revenue are often confused for one another. Be careful and note that revenue is the money “in the register” and profit is money after all costs are paid.
MEDIA LESSON Quadratic Applications – Cost (Duration 7:03 ) (Turn up the volume)
View the video lesson, take notes and complete the problems below
Example:
The cost C of producing 𝒂𝒂 “Total Cool Coolers” is modeled by the equation 𝑪𝑪 = 𝟎𝟎.𝟎𝟎𝟎𝟎𝟎𝟎𝒂𝒂𝟐𝟐 − 𝟎𝟎.𝟑𝟑𝒂𝒂 + 𝟏𝟏𝟏𝟏. How many coolers are produced when the cost is $19? (Round to the nearest whole number.)
MEDIA LESSON Quadratic Applications –Revenue (Duration 7:03) (Turn up the volume)
View the video lesson, take notes and complete the problems below
Example: The revenue, R, of producing and selling 𝑥𝑥 “Awesome Hearing Aids” is modeled by the equation 𝑹𝑹 = −𝟒𝟒𝒂𝒂𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝒂𝒂. How many hearing aids need to be produced and sold in order to earn a revenue of $850? You may have more than one answer. (Round to the nearest whole number.) YOU TRY
a) The cost, C, of producing 𝑥𝑥 “Totally Cool Coolers” is modeled by the equation
𝐶𝐶 = 0.005𝑥𝑥2 − 0.5𝑥𝑥 + 21 How many coolers are produced when the cost is $22? (Round to the nearest whole number.)
b) The Revenue, R, of producing and selling x Awesome Hearing Aids is modeled by the equation
𝑅𝑅 = −3𝑥𝑥2 + 87𝑥𝑥 How many hearing aids need to be produced and sold to earn a revenue of $611? You may have more than one answer. (Round to the nearest whole number.)
EXERCISE 1) Find the length of the missing leg given the
hypotenuse is 17 cm and the other leg is 12 cm. (Round to two decimal places.)
2) Find the length of the missing leg given the hypotenuse is 15 cm and the other leg is 12 cm. (Round to two decimal places.)
3) Find the length of the missing leg given the
hypotenuse is 8 cm and the other leg is 6 cm. (Round to two decimal places.)
4) Find the length of the hypotenuse if the length of the legs are 14 inches and 10 inches. Round to two decimal places.
5) Find the length of the hypotenuse if the length
of the legs are 10 inches and 12 inches. Round to two decimal places.
6) Find the length of the hypotenuse if the length of the legs are 14 inches and 12 inches. Round to two decimal places.
7) The cost, 𝐶𝐶 , of producing 𝑥𝑥 Totally Cool Coolers is modeled by the equation 𝐶𝐶 = 0.005𝑥𝑥2 − 0.3𝑥𝑥 + 18
How many coolers are produced when the cost is $19? (Round to the nearest whole number.)
8) The cost, 𝐶𝐶 , of producing 𝑥𝑥 Totally Cool Coolers is modeled by the equation
𝐶𝐶 = 0.005𝑥𝑥2 − 0.25𝑥𝑥 + 10 How many coolers are produced when the cost is $19? (Round to the nearest whole number.)
9) The revenue, 𝑅𝑅 , of producing and selling 𝑥𝑥 Awesome Hearing Aids is modeled by the equation
𝑅𝑅 = −5𝑥𝑥2 + 105𝑥𝑥 How many hearing aids need to be produced and sold to earn a revenue of $531? You may have more than one answer. (Round to the nearest whole number.)
10) The revenue, 𝑅𝑅 , of producing and selling 𝑥𝑥 Awesome Hearing Aids is modeled by the equation
𝑅𝑅 = −4𝑥𝑥2 + 120𝑥𝑥 How many hearing aids need to be produced and sold to earn a revenue of $880? You may have more than one answer. (Round to the nearest whole number.)
11) A rocket is launched at 𝑡𝑡 = 0 seconds. Its height, in meters above sea-level, is given by the equation
ℎ = −4.9𝑡𝑡2 + 325𝑡𝑡 + 227 After how many seconds does the rocket hit the ground? (Round to two decimal places.)
12) A rocket is launched at 𝑡𝑡 = 0 seconds. Its height, in meters above sea-level, is given by the equation ℎ = −4.9𝑡𝑡2 + 148𝑡𝑡 + 374
After how many seconds does the rocket hit the ground? (Round to two decimal places.)
Chapter 13
377
CHAPTER REVIEW KEY TERMS AND CONCEPTS
Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson.
