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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 β 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
Chapter 13
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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 arenβt 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 ππππππ + ππππ + ππ = ππ
Where ππππππ is called the leading term, ππππ is call the linear term, and ππ is called the constant coefficient (or constant term). Additionally, ππ β ππ.
SECTION 13.1: THE SQUARE ROOT PROPERTY A. SOLVE BASIC QUADRATIC EQUATIONS USING SQUAREROOT PROPERTY
Square root property
Let ππ β₯ ππ and ππ β₯ ππ. Then
ππππ = ππ if and only if ππ = Β±βππ In other words,
ππππ = ππ if and only if ππ = βππ or ππ = ββππ
MEDIA LESSON Solve basic quadratic equations using square root property (Duration 2:53)
View the video lesson, take notes and complete the problems below
Example: a) 8π₯π₯2 = 648 b) π₯π₯2 = 75
YOU TRY
Solve. a) π₯π₯2 = 81
b) π₯π₯2 = 44
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MEDIA LESSON Solve equations with even exponents (Duration 4:26)
View the video lesson, take notes and complete the problems below
Consider: 52= ________________ and (β5)2 = ________________________
When we clear an even exponent, we have ________________________________________________.
Example: Solve.
a) (5π₯π₯ β 1)2 = 49 b) οΏ½(3π₯π₯ + 2)44 = 81
YOU TRY
Solve. a) (π₯π₯ + 4)2 = 25
b) (6π₯π₯ β 9)2 = 45
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B. ISOLATE THE SQUARED TERM Letβs 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 β 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 β 2(2π₯π₯ + 1)2 = β46 b) 5(3π₯π₯ β 2)2 + 6 = 46
YOU TRY
Solve.
a) 5(3x β 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 π₯π₯2 = β1 does not have a real solution. There is a complex solution for this equation but we will not discuss it in this class.
Example: Solve
2ππ2 + 5 = 4 2ππ2 = 4 β 5 2ππ2 = β1
ππ2 = β12
This equation does not have a real solution.
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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 π₯π₯ term in the middle because if we apply the square root property to the π₯π₯ 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 π₯π₯ term in the middle and then apply the square root property to solve the equations. Recall: Perfect square formula ππππ + ππππππ+ ππππ = (ππ + ππ)ππ or ππππ β ππππππ + ππππ = (ππ β ππ)ππ
MEDIA LESSON Solve equations using square root property β Perfect Square formula (Duration 4:09)
View the video lesson, take notes and complete the problems below Example:
a) π₯π₯2 + 8π₯π₯ + 16 = 4 b) 9π₯π₯2 β 12π₯π₯ + 4 = 25
YOU TRY
Solve.
a) π₯π₯2 β 6π₯π₯ + 9 = 81
b) 9π₯π₯2 + 30π₯π₯ + 25 = 4
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EXERCISE Solve by applying the square root property.
1) (π₯π₯ β 3)2 = 16
2) (π₯π₯ β 2)2 = 49
3) (π₯π₯ β 7)2 = 4
4) (π π β 5)2 = 16
5) (ππ + 5)2 = 81
6) (π π + 3)2 = 4
7) (π‘π‘ + 9)2 = 37
8) (ππ + 5)2 = 57
9) (ππ β 9)2 = 63
10) (ππ + 1)2 = 125
11) (9ππ + 1)2 = 9
12) (7ππ β 8)2 = 36
13) (3π π β 6)2 = 25
14) 5(ππ β 7)2 β 6 = 369
15) 5(ππ β 5)2 + 13 = 103
16) 2ππ2 + 7 = 5
17) (2π π + 1)2 = 0
18) (π§π§ β 4)2 = 25
19) 3ππ2 + 2ππ = 2ππ + 24
20) 8ππ2 β 29 = 25 + 2ππ2
21) 2(ππ + 9)2 β 19 = 37
22) 3(ππ β 3)2 + 2 = 164
23) 7(2π₯π₯ + 6)2 β 5 = 170
24) 6(4π₯π₯ β 4)2 β 5 = 145
Apply the perfect square formula and solve the equations by using the square root property.
25) π₯π₯2 + 12π₯π₯ + 36 = 49
26) π₯π₯2 + 6π₯π₯ + 9 = 2
27) 16π₯π₯2 β 40π₯π₯ + 25 = 16
28) π₯π₯2 + 4π₯π₯ + 4 = 1
29) π₯π₯2 β 14π₯π₯ + 49 = 9
30) 25π₯π₯2 + 10π₯π₯ + 1 = 49
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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 π₯π₯2 β 2π₯π₯ β 7 = 0. This equation is not factorable, but there are two solutions to this equation: 1 + β2 and 1 β β2. 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:
(ππ + ππ)ππ = ππππ + ππππππ + ππππ ππππ (ππ β ππ)ππ = ππππ β ππππππ + ππππ
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 ππ.
ππππ + ππππππ + ππππ is easily factored to ________________________________
To make ππππ + ππππ+ ππ a perfect square, ππ = ___________________
Example:
a) π₯π₯2 + 10π₯π₯ + ππ
b) π₯π₯2 β 7π₯π₯ + ππ
c) π₯π₯2 β 37π₯π₯ + ππ d) π₯π₯2 + 6
5π₯π₯ + ππ
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Note
To complete the square of any trinomial, we always square half of the linear termβs coefficient, i.e.,
οΏ½πππποΏ½ππ
ππππ οΏ½πππππποΏ½
ππ
We usually use the second expression when the middle termβs coefficient is a fraction.
YOU TRY
Complete the square by finding ππ:
a) π₯π₯2 + 8π₯π₯ + ππ
b) π₯π₯2 β 7π₯π₯ + ππ
c) π₯π₯2 + 53π₯π₯ + ππ
B. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, a = 1
Steps to solving quadratic equations by completing the square
Given a quadratic equation ππππππ + ππππ + ππ = ππ, we can use the following method to solve for ππ.
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. ππππππ + ππππ __________ = ππ + _____________
Step 2. If ππ β ππ, divide both sides of the equation by ππ
Step 3. Complete the square, i.e., οΏ½πππποΏ½ππππππ οΏ½ππ
πππποΏ½
ππ 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.
(π₯π₯ β 5)2 = 28
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Example: a) π₯π₯2 + 6π₯π₯ β 9 = 0 b) π₯π₯2 β 5π₯π₯ + 10 = 0
c) 3π₯π₯2 + 2π₯π₯ β 9 = 0
YOU TRY
Solve.
a) π₯π₯2 + 10π₯π₯ + 24 = 0
b) ππ2 β 8ππ + 4 = 0
C. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, A β 1
MEDIA LESSON Solve quadratic equation by completing the square β a β 1 (Duration 4:59)
View the video lesson, take notes and complete the problems below To complete the square: πππ₯π₯2 + πππ₯π₯ + ππ = 0
1. Separate ____________________ and ______________________
2. Divide by ________________ (everything!)
3. Find _________ and _____________ to ______________________
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Example
a) 3π₯π₯2 β 15π₯π₯ + 18 = 0 b) 8π₯π₯ + 32 = 4π₯π₯2
YOU TRY
Solve.
a) 3π₯π₯2 β 36π₯π₯ + 60 = 0
b) 2ππ2 + ππ β 2 = 0
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EXERCISE Complete the square.
1) π₯π₯2 β 30π₯π₯ + _____
2) ππ2 β 36ππ + _____
3) π₯π₯2 β 15π₯π₯ + _____
4) π¦π¦2 β π¦π¦ + _____
5) ππ2 β 24ππ + _____
6) π₯π₯2 β 34π₯π₯ + _____
7) ππ2 β 19ππ + _____
8) ππ2 β 17ππ + _____
Solve each equation by completing the square. If the solution is not a real solution, then state not a real solution.
9) 6ππ2 + 12ππ β 24 = β6
10) 6ππ2 β 12ππ β 14 = 4
11) ππ2 β 56 = 10ππ
12) π₯π₯2 + 8π₯π₯ + 15 = 8
13) 8ππ2 + 16ππ = 64
14) ππ2 + 4ππ = 12
15) π₯π₯2 β 16π₯π₯ + 55 = 0
16) ππ2 = β21 + 10ππ
17) 4ππ2 β 15ππ + 56 = 3ππ2
18) ππ2 + 7ππ β 33 = 0
19) π₯π₯2 + 10π₯π₯ β 57 = 4
20) ππ2 β 8ππ β 12 = 0
21) ππ2 β 16ππ + 67 = 4
22) π₯π₯2 = β10π₯π₯ β 29
23) 5ππ2 β 10ππ + 48 = 0
24) 7ππ2 β ππ + 7 = 7ππ + 6ππ2
25) 2π₯π₯2 + 4π₯π₯ + 38 = β6 26) 8ππ2 + 16ππ β 37 = 5
27) 5π₯π₯2 + 5π₯π₯ = β31β 5π₯π₯ 28) π£π£2 + 5π£π£ + 28 = 0
29) ππ2 β 7ππ + 50 = 3
30) 5π₯π₯2 + 8π₯π₯ β 40 = 8
31) 8ππ2 + 10ππ = β55
32) β2π₯π₯2 + 3π₯π₯ β 5 = β4π₯π₯2
33) 8ππ2 + 16ππ β 1 = 0
34) ππ2 β 16ππ β 52 = 0
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SECTION 13.3: QUADRATIC FORMULA The quadratic formula is derived from the method of completing the square. If we took a general quadratic equation
πππ₯π₯2 + πππ₯π₯ + ππ = 0
and solved for π₯π₯ by completing the square, we would obtain the quadratic formula. Letβs try this.
MEDIA LESSON Deriving the Quadratic Formula (Duration 4:04)
Quadratic formula
Given the quadratic equation πππ₯π₯2 + πππ₯π₯ + ππ = 0. Then
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
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 π₯π₯.
Discriminant
Given the quadratic equation πππ₯π₯2 + πππ₯π₯ + ππ = 0, then its discriminant, denoted as the upper-case
Greek letter delta, Ξ, is defined as π«π« = ππππ β ππππππ
βπ«π« = οΏ½ππππ β ππππππ
Quadratic formula:
ππ =βππ Β± βπ«π«
ππππ
Case 1: If Ξ is positive, the equation has 2 solutions.
Case 2: If Ξ is zero, the equation has 1 solution.
Case 3: If Ξ is negative, the equation has no real solution. (The equation has two complex solutions
but we will not discuss complex numbers in this class).
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MEDIA LESSON Determinant (Duration 4:58)
View the video lesson, take notes and complete the problems below
πππ₯π₯2 + πππ₯π₯ + ππ = 0 π₯π₯ = _______________________________ β’ If ππ2 β 4ππππ ________________________________________________________________________
β’ If ππ2 β 4ππππ ________________________________________________________________________
β’ If ππ2 β 4ππππ ________________________________________________________________________
Example: Determine the number of solutions to the quadratic equation.
π₯π₯2 + 14π₯π₯ + 49 = 0
MEDIA LESSON Determinant examples (Duration 4:58)
View the video lesson, take notes and complete the problems below
Example: Describe the type of solutions to the quadratic equation.
a) π₯π₯2 β 3π₯π₯ β 28 = 0 ππ = ______ ππ = ______ ππ =______
a) π₯π₯2 β 4π₯π₯ + 12 = 0 ππ = ______ ππ = ______ ππ =______
b) β2π₯π₯2 + π₯π₯ + 5 = 0 ππ = ______ ππ = ______ ππ =______
c) 2π₯π₯2 + 8π₯π₯ + 8 = 0 ππ = ______ ππ = ______ ππ =______
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YOU TRY
Find the determinant and determine how many solutions each of the following equation has.
a) π₯π₯2 + 3π₯π₯ β 4 = 0
b) 2π₯π₯2 + 4π₯π₯ + 1 = 0
c) π₯π₯2 β 4π₯π₯ + 4 = 0
d) 8ππ2 + 5ππ + 1 = 0
B. APPLY THE QUADRATIC FORMULA
MEDIA LESSON Solve equations uing quadratic formula β two real rational solutions (Duration 3:14 )
View the video lesson, take notes and complete the problems below
Example: Solve using the quadratic formula.
6π₯π₯2 β π₯π₯ β 15 = 0
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MEDIA LESSON Solve equations uing quadratic formula β two real irrational solutions (Duration 4:16 )
View the video lesson, take notes and complete the problems below
Example: Solve using the quadratic formula.
2π₯π₯2 β 4π₯π₯ β 3 = 0
MEDIA LESSON Solve equations uing quadratic formula β one solution (Duration 2:50 )
View the video lesson, take notes and complete the problems below
Example: Solve using the quadratic formula.
4π₯π₯2 β 12π₯π₯ + 9 = 0 YOU TRY
Solve.
a) π₯π₯2 + 3π₯π₯ + 2 = 0
b) 3π₯π₯2 + 2π₯π₯ β 7 = 0
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C. MAKE ONE SIDE OF AN EQUATION EQUAL TO ZERO
MEDIA LESSON Set one side of an equation equal to 0 before solving (Duration 4:46)
View the video lesson, take notes and complete the problems below
Before using the quadratic formula, the equation must equal ________. Example:
a) 2π₯π₯2 = 15 β 7π₯π₯ b) 3π₯π₯2 β 5π₯π₯ + 2 = 7
YOU TRY
Solve.
a) 25π₯π₯2 = 30π₯π₯ + 11
b) 3π₯π₯2 + 4π₯π₯ + 8 = 2π₯π₯2 + 6π₯π₯ β 5
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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) 2π₯π₯2 β 8π₯π₯ β 2 = 0 2) 3ππ2 β 2ππ β 1 = 0
3) 2π₯π₯2 + 5π₯π₯ = β3
4) π£π£2 β 4π£π£ β 5 = β8
5) 2ππ2 + 3ππ + 14 = 6
6) 3ππ2 + 3ππ β 4 = 7
7) 7π₯π₯2 + 3π₯π₯ β 16 = β2
8) 2ππ2 + 6ππ β 16 = 4
9) 3ππ2 + 3ππ = β3
10) 2π₯π₯2 = β7π₯π₯ + 49
11) 5π₯π₯2 = 7π₯π₯ + 7
12) 8ππ2 = β3ππ β 8
13) 4ππ2 + 5ππ β 36 = 3ππ2
14) β5ππ2 β 3ππ β 52 = 2 β 7ππ2
15) 7ππ2 β 12 = β3ππ
16) 2π₯π₯2 + 4π₯π₯ + 12 = 8
17) 5ππ2 + 2ππ + 6 = 0
18) 2π₯π₯2 β 2π₯π₯ β 15 = 0
19) 6ππ2 β 3ππ + 3 = β4
20) ππ2 + 4ππ β 48 = β3
21) 4ππ2 + 5ππ = 7
22) 3ππ2 + 4 = β6ππ
23) 3ππ2 β 3 = 8ππ
24) 6π£π£2 = 4 + 6π£π£
25) 6ππ2 = β5ππ + 13
26) 2ππ2 + 6ππ β 16 = 2ππ
27) 6ππ2 = ππ2 + 7 β ππ
28) 49ππ2 β 28ππ + 4 = 0
29) 12π₯π₯2 + π₯π₯ + 7 = 5π₯π₯2 + 5π₯π₯
30) π₯π₯2 β 6π₯π₯ + 9 = 0
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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 triangleβs legs is the same as the square of the length of the triangleβs hypotenuse. This property- which has many applications in science, art, engineering, and architecture β is now called the Pythagorean Theorem.
Pythagorean Theorem
If ππ and ππ are the lengths of the legs of a right triangle and ππ 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:
(ππππππππππ ππππππ)ππ + (ππππππππππππ ππππππ)ππ = (ππππππππππππππππππππ)ππ
ππππ + ππππ = ππππ
MEDIA LESSON Pythagorean theorem applications (Duration 5:00)
View the video lesson, take notes and complete the problems below
Pythagorean Theorem- Find the hypotenuse.
Name the sides of the right triangle:
Pythagorean Theorem:
_____________________________
ππ is always the _______________________
Example: a) Find the missing side.
b) The base of a ladder is four feet from a building. The top of the ladder is eight feet up the building. How long is the ladder?
Chapter 13
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YOU TRY
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 β 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 π‘π‘ seconds after taking off is given by the formula β = β3π‘π‘2 + 14π‘π‘ + 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.
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YOU TRY
a) A rocket is launched at π‘π‘ = 0 seconds. Its height, in meters above sea-level, is given by the equation β = β4.9π‘π‘2 + 52π‘π‘ + 367
At what time does the rocket hit the ground? (Round answer to 2 decimal places.) Note: With applications, such as these, it doesnβt 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 βin the registerβ and profit is money after all costs are paid.
MEDIA LESSON Quadratic Applications β 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 ππ βTotal Cool Coolersβ is modeled by the equation πͺπͺ = ππ.ππππππππππ β ππ.ππππ + ππππ. How many coolers are produced when the cost is $19? (Round to the nearest whole number.)
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MEDIA LESSON Quadratic Applications βRevenue (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 π₯π₯ βAwesome Hearing Aidsβ is modeled by the equation πΉπΉ = βππππππ + ππππππππ. 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 π₯π₯ βTotally Cool Coolersβ is modeled by the equation
πΆπΆ = 0.005π₯π₯2 β 0.5π₯π₯ + 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
π π = β3π₯π₯2 + 87π₯π₯ 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.)
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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, πΆπΆ , of producing π₯π₯ Totally Cool Coolers is modeled by the equation πΆπΆ = 0.005π₯π₯2 β 0.3π₯π₯ + 18
How many coolers are produced when the cost is $19? (Round to the nearest whole number.)
8) The cost, πΆπΆ , of producing π₯π₯ Totally Cool Coolers is modeled by the equation
πΆπΆ = 0.005π₯π₯2 β 0.25π₯π₯ + 10 How many coolers are produced when the cost is $19? (Round to the nearest whole number.)
9) The revenue, π π , of producing and selling π₯π₯ Awesome Hearing Aids is modeled by the equation
π π = β5π₯π₯2 + 105π₯π₯ 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, π π , of producing and selling π₯π₯ Awesome Hearing Aids is modeled by the equation
π π = β4π₯π₯2 + 120π₯π₯ 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 π‘π‘ = 0 seconds. Its height, in meters above sea-level, is given by the equation
β = β4.9π‘π‘2 + 325π‘π‘ + 227 After how many seconds does the rocket hit the ground? (Round to two decimal places.)
12) A rocket is launched at π‘π‘ = 0 seconds. Its height, in meters above sea-level, is given by the equation β = β4.9π‘π‘2 + 148π‘π‘ + 374
After how many seconds does the rocket hit the ground? (Round to two decimal places.)
Chapter 13
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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.
Quadratic equation
Square root property
Discriminant
Quadratic formula
Pythagorean Theorem
Cost and Revenue
