1
Lesson 7.1: Solving Quadratic Equations by Graphing, Factoring, and
Square Root Method
Learning Goals:
1) How do we solve a quadratic equation by graphing?
2) How do we solve a quadratic equation by factoring?
3) How do we solve a quadratic equation by the square root method?
Warm up:
1) Solve the given equation 3 different ways. π₯2 β 4 = 0
Graphically Factoring Square Root Method
DOPS (π₯ β 2)(π₯ + 2) = 0 π₯ β 2 = 0 π₯ + 2 = 0
π₯ = Β±2
π₯2 = 4
βπ₯2 = β4 π₯ = Β±2
2) Express in simplest radical form:
ββ25
ββ1 β β25 5π
ββ32
ββ16 β β2
4πβ2
β4
5
2 β β5
β5 β β5
2β5
5
Advice to solving Quadratic Equations:
Express quadratic in standard form ππ₯2 + ππ₯ + π = 0
Determine appropriate methods to solve quadratic equations.
2
1. Solve the following equation by factoring: π₯2 β 48 = 2π₯
Set = 0
π₯2 β 2π₯ β 48 = 0
(π₯ β 8)(π₯ + 6) = 0
π₯ β 8 = 0 π₯ + 6 = 0
π₯ = 8 & β 6
2. Solve for π₯ and express in simplest form: 3π₯2 + 9 = 0
3π₯2 = β9
π₯2 = β3
π₯ = Β±πβ3
3. Use graphing to find the roots of π₯2 β 2π₯ = 8
1. Solve for π: π2 = 5π
π2 β 5π = 0 set = 0
π(π β 5) = 0 GCF = π
π = 0 π β 5 = 0
π = 0 & 5
3
2. Find the zeros of π(π₯) = 5π₯2 β 8π₯ β 4
0 = 5π₯2 β 8π₯ β 4 AC method
0 = 5π₯2 β 10π₯ + 2π₯ β 4 5(β4) = β20
0 = 5π₯(π₯ β 2) + 2(π₯ β 2) β10, 2
0 = (5π₯ + 2)(π₯ β 2)
π₯ = β2
5 & 2
3. Solve for π₯: (π₯ + 3)2 = 16
β(π₯ + 3)2 = β16
π₯ + 3 = Β±4
π₯ = β3 Β± 4
β3 + 4 β 3 β 4
π₯ = 1 & β 7
4. Algebraically solve for π₯ in simplest form: 3π₯2 + 4 = 0
3π₯2 = β4
π₯2 = β4
3
π₯ = Β±ββ4
β3= Β±
2π
β3 ββ3
β3 or Β±
2πβ3
3
5. Find the roots of the equation: 2π₯2 + π₯ β 3 = (π₯ β 1)(π₯ + 2)
2π₯2 + π₯ β 3 = π₯2 + 2π₯ β π₯ β 2
2π₯2 + π₯ β 3 = π₯2 + π₯ β 2
π₯2 β 1 = 0
π₯ = Β±1
4
6. Find the zeros of π₯ β 4 =5
π₯ Cross Multiply!
π₯(π₯ β 4) = 5
π₯2 β 4π₯ = 5
π₯2 β 4π₯ β 5 = 0
(π₯ β 5)(π₯ + 1) = 0
π₯ = 5 & β 1
7. Solve for π₯: 1
4(π₯ β 6)2 = 8
(π₯ β 6)2 = 32
π₯ β 6 = Β±β32
π₯ = 6 Β± 4β2
8. Solve for π₯: (π₯ + 1)2 β 87 = π₯2
(π₯ + 1)(π₯ + 1) β 87 = π₯2
π₯2 + 2π₯ + 1 β 87 = π₯2
2π₯ β 86 = 0
2π₯ = 86
π₯ = 43
5
Check for understandingβ¦
1. On a test, Samantha says that the roots of the equation π₯2 + 2π₯ β 3 = 0 are
β1 and 3. Before she hands in her test, she wants to check over her work.
Show the check that Samantha could do on her test and state whether or not she
is correct. If she is not correct, give the correct solution.
(β1)2 + 2(β1) β 3 = 0 (3)2 + 2(3) β 3 = 0
1 β 2 β 3 = 0 9 + 6 β 3 = 0
β4 β 0 12 β 0
Therefore she is not correct!
(π₯ + 3)(π₯ β 1) = 0
π₯ = β3 & 1
2. Philβs teacher gave the class the quadratic function π(π₯) = (π₯ β 2)2 β 4.
a) State two different methods Phil could use to solve the equation π(π₯) = 0.
b) Using one of the methods stated in part a, solve π(π₯) = 0 for π₯.
Factoring, Quadratic Formula
π(π₯) = (π₯ β 2)(π₯ β 2) β 4
π(π₯) = π₯2 β 4π₯ + 4 β 4
π(π₯) = π₯(π₯ β 4)
π₯ = 0 & 4
6
Homework Lesson 7.1: Solving Quadratic Equations by Graphing,
Factoring, and Square Root Method
1. Solve for π₯: π₯ β 4 =β3
π₯
2. Find the roots of 3π₯2 β 5π₯ = 36 β 2π₯
3. Solve for π₯ in simplest form: 3π₯2 + 25 = 0
4. Solve for π₯: 5(π₯ β 1)2 = 50
7
5. Use graphing to solve the given
equation for π₯: π₯2 + 5π₯ + 6 = 0
6. Find the zeros of π(π₯) = 4π₯2 β 16 β 20
7. Solve for π₯ in simplest form: (2π₯ + 3)(π₯ β 4) = 2π₯2 + 13π₯ + 15
8. Solve for π₯: π₯2 + π₯ β 1 = (β4π₯ + 3)2
8
Lesson 7.2: Solving Quadratics by Completing the Square
Learning Goal: How do we solve a quadratic equation by completing the
square?
Practice: Find the value of π that makes the expression a perfect square
trinomial. Then write the expression as the square of a binomial.
π₯2 + 14π₯ + π
π = (14
2)2
= 72 = 49
π₯2 + 14π₯ + 49 (π₯ + 7)(π₯ + 7) (π₯ + 7)2
π₯2 β 9π₯ + π
π = (β9
2)2
=81
4
π₯2 β 9π₯ +81
4
(π₯ β9
2)2
Steps to Completing the Square
1. The "π" coefficient must equal 1 (divide all terms by "π").
2. Isolate the π₯-terms (move "π" to the other side).
3. Take (π
2)2 and add that number to each side.
4. Factor the perfect square trinomial and express it as (π₯ + π)2.
5. Solve for π₯ by using the square root method.
9
Completing the Square when π = 1
Example 1: Solve π₯2 β 10π₯ + 13 = 0 by completing the square.
Solution Steps
π₯2 β 10π₯ + 13 = 0 Write original equation.
π₯2 β 10π₯ = β13 Write left side in the form π₯2 + ππ₯, I say βisolate the π₯β
π₯2 β 10π₯ + π = β13 + π
π = (β10
2)
2
= (β5)2 = 25
Complete the square, I say βdivide by two and square itβ
π₯2 β 10π₯ + 25 = β13 + 25
(π₯ β 5)2 = 12
Write left side as a binomial squared.
β(π₯ β 5)2 = β12 Take the square roots of each side.
π₯ β 5 = Β±β12
π₯ = 5 Β± β12
Solve for π₯.
π₯ = 5 Β± 2β3 Simplify
Practice: Solve π₯(π₯ + 3) = β2 by completing the square, and express the results
in simplest form.
π₯2 + 3π₯ = β2
π₯2 + 3π₯ + π = β2 + π
π = (3
2)2=9
4
π₯2 + 3π₯ +9
4= β2 +
9
4
(π₯ +3
2)2=1
4
π₯ +3
2= Β±
1
2
π₯ = β3
2Β±1
2
π₯ = β1 & β 2
10
Completing the Square when π β 1
Example 1: Solve 3π₯2 β 12π₯ + 27 = 0 by completing the square.
Solution Steps
3π₯2 β 12π₯ + 27 = 0 Write original equation.
π₯2 β 4π₯ + 9 = 0 Get π = 1before you can complete the square! Divide each side by the coefficient
of π₯2. π₯2 β 4π₯ = β9 Write left side in the form π₯2 + ππ₯.
π₯2 β 4π₯ + π = β9 + π
π = (β4
2)2
= (β2)2 = 4
Complete the square.
π₯2 β 4π₯ + 4 = β9 + 4
(π₯ β 2)2 = β5
Write left side as a binomial squared.
β(π₯ β 2)2 = ββ5 Take the square roots of each side.
π₯ β 2 = Β±ββ5
π₯ = 2 Β± ββ5
Solve for π₯.
π₯ = 2 Β± πβ5 Simplify
Practice: Solve 4π₯2 β 20π₯ = β9 by completing the square, and express the
results in simplest form.
π₯2 β 5π₯ = β9
4 Isolate the π₯ and make sure π = 1
π₯2 β 5π₯ + π = β9
4+ π
π = (β5
2)2=25
4
π₯2 β 5π₯ +25
4= β
9
4+25
4
(π₯ β5
2)2= 4
π₯ β5
2= Β±2
π₯ =5
2Β± 2
π₯ =9
2 &
1
2
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Practice: Solve the following equations by completing the square, and express
the results in simplest form.
a) π₯2 β 9π₯ β 1 = 0 b) 5π₯(π₯ + 6) = β50
π₯2 β 9π₯ = 1 5π₯2 + 30π₯ = β50
π₯2 β 9π₯ + π = 1 + π π₯2 + 6π₯ = β10
π = (β9
4)2=81
16 π₯2 + 6π₯ + π = β10 + π
π₯2 β 9π₯ +81
16= 1 +
81
16 π = (
6
2)2= (3)2 = 9
(π₯ β9
4)2=97
16 π₯2 + 6π₯ + 9 = β10 + 9
π₯ β9
4= Β±
β97
4 (π₯ + 3)2 = β1
π₯ =9
4Β±β97
4 π₯ + 3 = Β±π
π₯ = β3 Β± π
Analysis Question: Which equation has the same solutions as π₯2 + 6π₯ β 7 = 0?
1. (π₯ + 3)2 = 2 2. (π₯ β 3)2 = 2 3. (π₯ β 3)2 = 16 4. (π₯ + 3)2 = 16
π₯2 + 6π₯ = 7
π₯2 + 6π₯ + π = 7 + π
π = (6
2)2= (3)2 = 9
π₯2 + 6π₯ + 9 = 7 + 9
(π₯ + 3)2 = 16
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Homework 7.2: Solving Quadratics by Completing the Square
1. Find the value of π that makes the expression a perfect square trinomial.
Then write the expression as the square of a binomial.
π₯2 + 10π₯ + π
π¦2 β 7π¦ + π
2. Solve the following equations by completing the square and express the
results in simplest form.
a. π₯2 + 6π₯ + 3 = 0 b. π‘(π‘ β 8) = 5
c. 7π¦2 + 28π¦ + 56 = 0 d. 4π₯2 β 30π₯ = β9 β 10π₯
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3. Determine whether you would use factoring or the square root method to
solve each equation. Then solve the equation.
a. π₯2 β 4π₯ β 21 = 0 b. (π₯ + 4)2 = 16
c. 5π₯2 β 10π₯ + 1 = 2π₯2 β 5π₯ + 13 d. (π₯ + 3)2 = 17π₯ + 21
4. Error Analysis: Describe and correct the error in solving the equation below
4π₯2 + 24π₯ β 11 = 0
4(π₯2 + 6π₯) = 11
4(π₯2 + 6π₯ + 9) = 11 + 9
4(π₯ + 3)2 = 20
(π₯ + 3)2 = 5
π₯ + 3 = Β±β5
π₯ = β3 Β± β5
14
5. Determine the roots of the following equations graphically:
a. π₯2 β 5π₯ β 14 = 0 b. π₯2 = 9
6. The zeros of the function π(π₯) = (π₯ + 2)2 β 25 are
1) -2 and 5 2) -3 and 7 3) -5 and 2 4) -7 and 3
7. A student was given the equation π₯2 + 6π₯ β 13 = 0 to solve by completing the
square. The first step that was written is shown below.
π₯2 + 6π₯ = 13
The next step in the studentβs process was π₯2 + 6π₯ + π = 13 + π. State the value
of π that creates a perfect square trinomial. Explain how the value of π is
determined.
15
Lesson 7.3: Solving Quadratics by Using the Quadratic Formula
Learning Goals:
1) What is the quadratic formula?
2) How do we solve a quadratic equation by using the quadratic formula?
Do Now: Solve for π₯ by completing the square and express your answer in
simplest radical form: 2π₯2 + 2 = 6π₯
2π₯2 β 6π₯ + 2 = 0
π₯2 β 3π₯ + 1 = 0 Get π = 1
π₯2 β 3π₯ = β1
π₯2 β 3π₯ + π = β1 + π
π = (β3
2)2=9
4
π₯2 β 3π₯ +9
4= β1 +
9
4
(π₯ β3
2)2=5
4
π₯ β3
2= Β±
β5
2
π₯ =3
2Β±β5
2
The Quadratic Formula
Previously, you solved quadratic equations by completing the square.
In this lesson, you will learn about solving quadratic equations by using a
formula that is derived by completing the square for the general equation
ππ₯2 + ππ₯ + π = 0.
The formula is called the Quadratic Formula. π₯ =βπΒ±βπ2β4ππ
2π On
Reference Sheet
16
Example: Solve for π₯ by using the quadratic formula and express your answer in
simplest radical form: 2π₯2 + 2 = 6π₯
2π₯2 β 6π₯ + 2 = 0 Set = 0
π = 2, π = β6, π = 2
π₯ =βπΒ±βπ2β4ππ
2π=β(β6)Β±β(β6)2β4(2)(2)
2(2)
π₯ =6Β±β36β16
4=6Β±β20
4
π₯ =6Β±2β5
4=3Β±β5
2
Steps to Using the Quadratic Formula
1. Set quadratic equation equal to zero (ππ₯2 + ππ₯ + π = 0)
2. Identify the π, π, and π coefficients
3. Substitute π, π, and π into the quadratic formula
4. Simplify the formula carefully
5. Look to simplify and reduce (always start with radical first)
Summary of Quadratic Formula
What is the quadratic formula? π =βπΒ±βππβπππ
ππ
When should you use the quadratic formula? When you cannot solve by factoring
17
Model Problem: Solve for π: 2π2 β 8 = 3π
2π2 β 3π β 8 = 0
π = 2, π = β3, and π = β8
π =β(β3)Β±β(β3)2β4(2)(β8)
2(2)=3Β±β9+64
4
π =3Β±β73
4
Directions: Use the quadratic formula to solve each equation.
1. Solve for π: 1
3π2 + 2π + 8 = 5
1
3π2 + 2π + 3 = 0
3 (1
3π2 + 2π + 3 = 0) Easier to eliminate the fraction!
π2 + 6π + 9 = 0
π = 1, π = 6, and π = 9
π =β(6)Β±β(6)2β4(1)(9)
2(1)=β6Β±β36β36
2
π = β3
2. Solve for π₯: 5π₯ β 7π₯2 = 3π₯ + 4
β7π₯2 + 2π₯ β 4 = 0
β1(β7π₯2 + 2π₯ β 4 = 0) Easier to have a positive coefficient for π!
7π₯2 β 2π₯ + 4 = 0
π = 7, π = β2, and π = 4
π₯ =β(β2)Β±β(β2)2β4(7)(4)
2(7)=2Β±ββ108
14
π₯ =2Β±6πβ3
14=1Β±3πβ3
7
18
Application of the Quadratic Formula
Barb pulled the plug in her bathtub and it started to drain. The amount of water
in the bathtub as it drains is represented by the equation πΏ = β5π‘2 β 8π‘ + 120,
where πΏ represents the number of liters of water in the bathtub and π‘ represents
the amount of time, in minutes, since the plug was pulled.
How many liters of water were in the bathtub when Barb pulled the plug? Show
your reasoning. πΏ =? , π‘ = 0 because no time has passed!
πΏ = β5(0)2 β 8(0) + 120
πΏ = 120
Determine, to the nearest tenth of a minute, the amount of time it takes for all the
water in the bathtub to drain. π‘ =? , πΏ = 0 no more water!
0 = β5π‘2 β 8π‘ + 120
β1(0 = β5π‘2 β 8π‘ + 120)
0 = 5π‘2 + 8π‘ β 120
π = 5, π = 8, and π = β120
π‘ =β(8)Β±β(8)2β4(5)(β120)
2(5)=β8Β±β64+2400
10=β8Β±β2464
10
π‘ = 4.2 & β 5.8 (but omit because time canβ²t be negative)
19
Homework 7.3: Solving Quadratics by Using the Quadratic Formula
1. Solve the following equations by using the quadratic formula and express the
results in simplest form.
a. π₯2 + 2π₯ β 8 = 0 b. 2π¦2 + 3π¦ β 5 = 4
2. Mattβs rectangular patio measures 9 feet by 12 feet. He wants to increase the
patioβs dimensions so its area will be twice the area it is now. He plans to
increase both the length and the width by the same amount, π₯. Find π₯, to the
nearest hundredth of a foot.
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3. Determine the best method to solve each equation and use it to find all values
of π₯ in simplest form.
a. π₯2 β 2π₯ = 12
b. (π₯ β 2)2 = 8
c. 2π₯2 β 54 = 12π₯
d. 5π₯2 + 38 = 3
4. Find all real solutions to the equation (π₯2 β 6π₯ + 3)(2π₯2 β 4π₯ β 7) = 0
21
Lesson 7.4: Solving Systems of Equations Graphically
Learning Goals:
1) How can we change an equation of a circle from standard form to center-
radius form?
2) How can we solve a quadratic/linear system of equations graphically?
Warm-Up:
1. Suppose you were given an equation for a circle and an equation for a line.
What possibilities are there for the two figures to intersect? Sketch a graph for
each possibility.
22
2. Suppose you were given an equation for a parabola and an equation for a
line. What possibilities are there for the two figures to intersect? Sketch a graph
for each possibility.
3. State the center and radius of the given circles:
a) (π₯ β 1)2 + (π¦ β 2)2 = 25 Center: (1, 2) & π = β25 = 5
b) (π₯ + 5)2 + (π¦ β 2)2 = 29 Center: (β5, 2) & π = β29
c) π₯2 + (π¦ β 2)2 = 4 Center: (0, 2) & π = β4 = 2
4. Rewrite π₯2 + π¦2 β 4π₯ + 2π¦ = β1 by completing the square in both π₯ and π¦.
Describe the circle represented by this equation.
π₯2 β 4π₯β + π¦2 + 2π¦β = β1 (β4
2)2= (β2)2 = 4 (
2
2)2= 1
π₯2 β 4π₯ + 4 + π¦2 + 2π¦ + 1 = β1 + 4 + 1
(π₯ β 2)2 + (π¦ + 1)2 = 4 Center: (2,β1) and Radius= 2
23
5. The equation of a circle is π₯2 + π¦2 + 6π¦ = 7. What are the coordinates of the
center and the length of the radius of the circle?
π₯2 + π¦2 + 6π¦ = 7 (6
2)2= (3)2 = 9
π₯2 + π¦2 + 6π¦ + 9 = 7 + 9
π₯2 + (π¦ + 3)2 = 16 Center: (0,β3) and Radius= 4
6. If the equation of a circle is 2π₯2 + 2π¦2 β 32π₯ + 12π¦ + 52 = 0, find the length of
the radius and the coordinates of the center of the circle.
2π₯2 + 2π¦2 β 32π₯ + 12π¦ + 52 = 0
2
π₯2 + π¦2 β 16π₯ + 6π¦ + 26 = 0
π₯2 β 16π₯ + π¦2 + 6π¦ = β26 (β16
2)2= (β8)2 = 64 (
6
2)2= (3)2 = 9
π₯2 β 16π₯ + 64β + π¦2 + 6π¦ + 9β = β26 + 64 + 9
(π₯ β 8)2 + (π¦ + 3)2 = 47 Center: (8,β3) and Radius= β47
7. If the equation of a circle is 4π₯2 + 4π¦2 β 32π₯ + 8π¦ + 16 = 0, find the length of
the radius and the coordinates of the center of the circle.
4π₯2 + 4π¦2 β 32π₯ + 8π¦ + 16 = 0
4
π₯2 + π¦2 β 8π₯ + 2π¦ + 4 = 0
π₯2 β 8π₯ + π¦2 + 2π¦ = β4 (β8
2)2= (β4)2 = 16 (
2
2)2= (1)2 = 1
π₯2 β 8π₯ + 16β + π¦2 + 2π¦ + 1β = β4 + 16 + 1
(π₯ β 4)2 + (π¦ + 1)2 = 13 Center: (β4, 1) and Radius= β13
Remember to put the 4 back in: 4(π₯ β 4)2 + 4(π¦ + 1)2 = 52
24
Graphing Systems of Equations:
1. Graph the line given by 3π₯ + 4π¦ = 25 and the circle given by π₯2 + π¦2 = 25.
Find all solutions to the system of equations.
3π₯ + 4π¦ = 25
4π¦ = β3π₯ + 25
π¦ = β3
4π₯ +
25
4
π¦ = β3
4π₯ + 6.25
π₯2 + π¦2 = 25
Center: (0, 0) π = 5
Solution: (4, 3)
2. Graph the line given by π₯ +
π¦ = β2 and the quadratic curve
given by
π¦ = π₯2 β 4. Find all solutions to
the system of equations.
Solutions: (β2, 0) & (1,β3)
25
3. Find all solutions to the
following system of equations.
5π¦ β 5π₯ = 30
π₯2 + π¦2 + 4π₯ β 2π¦ β 4 = 0
π¦ = π₯ + 6
π₯2 + 4π₯ + π¦2 β 2π¦ = 4
(π₯ + 2)2 + (π¦ β 1)2 = 9
Center: (β2, 1) π = 3
Solutions: (β5, 1) & (β2, 4)
4. Find all values of π so that the following system has two solutions.
π₯2 + π¦2 = 25
π¦ = π
Center: (0, 0) π = 5
β5 < π < 5
26
Real-World Application:
An asteroid is moving in a parabolic
arc that is modeled by the function
π(π₯) = π₯2 β 4π₯ + 9 where π₯
represents time. A laser is on the
path of π(π₯) = 2π₯ + 4. When will
the laser first hit the asteroid?
(1) (0, 9) & (1, 6)
(2) (1, 6) & (5, 14)
(3) (1, 6)
(4) (2, 5)
27
Homework 7.4: Solving Systems of Equations Graphically
1. Which of the following systems of equations has exactly one point of
intersection?
(1) π¦ = π₯2 β 5π₯ + 7 and π¦ β 1 = 2π₯
(2) π¦ β 4π₯2 = π₯ β 3 and π¦ = 3
(3) π¦ β π₯2 = 2π₯ + 4 and π¦ = 3
(4) π¦ + 9π₯2 = β8 and π¦ = β1
2. Given the equation of a circle is 3π₯2 + 3π¦2 β 12π₯ + 30π¦ β 10 = 0, state the
length of the radius and coordinates of the center.
3. Find all values of π so that the
following system has exactly one solution.
Illustrate with a graph.
π¦ = 5 β (π₯ β 3)2
π¦ = π
28
4. Find all solutions to the following
system of equations.
π¦ + 2π₯ = 3
π¦ = π₯2 β 6π₯ + 3
5. Solve the following system of
equations graphically.
{2π₯ + π¦ = 15
(π₯ β 2)2 + (π¦ β 1)2 = 25
29
Lesson 7.5: Solving Systems of Equations Algebraically
Learning Goals:
1) How can we solve a quadratic/linear system of equations algebraically?
2) How is the solution to a quadratic/linear system related to its graphical
solution?
Warm-Up: Solve the system π₯2 + π¦2 = 9 and π₯ β 3π¦ = 3 graphically.
Center: (0, 0) & π = 3
β3π¦ = βπ₯ + 3
π¦ =1
3π₯ β 1
Solutions: (3, 0)& ?
What is difficult about solving this
graphically? Hard to draw a circle
without a compass
Steps to Solving a System of Equations Algebraically
1) Get one equation to be written as π¦ = or π₯ =
2) Substitute this equation into the other equation for π₯ or π¦
3) Simplify this equation and set it equal to zero.
4) Use one of the methods for solving a quadratic to solve for the variable
Factoring, square root method, completing the square, quadratic formula
5) Plug your answer back into one of the equations to solve for the other variable.
30
1. Find all solutions of the system of equations algebraically:
π₯2 + π¦2 = 9
π₯ β 3π¦ = 3
Substitute: π₯ = 3π¦ + 3
(3π¦ + 3)2 + π¦2 = 9 Now find π₯-values!
(3π¦ + 3)(3π¦ + 3) + π¦2 = 9 π₯ β 3(0) = 3 π₯ β 3 (β9
5) = 3
9π¦2 + 18π¦ + 9 + π¦2 = 9 π₯ β 0 = 3 π₯ +27
5= 3
10π¦2 + 18π¦ = 0 π₯ = 3 π₯ = β2.4 or β12
5
2π¦(5π¦ + 9) = 0
π¦ = 0 and y = β9
5 Solutions: (3, 0) and (β
12
5, β
9
5)
How is the solution to a system or equations related to its graphical solution?
2. Find all solutions of the system of equations algebraically:
π¦2 β 2π₯2 = 6
π¦ = β2π₯
Substitute: π¦ = β2π₯
(β2π₯)2 β 2π₯2 = 6 Now find π¦-values!
4π₯2 β 2π₯2 = 6 π¦ = β2(β3) π¦ = β2(ββ3)
2π₯2 = 6 π¦ = β2β3 π¦ = 6β3
π₯2 = 3
π₯ = Β±β3 Solutions: (β3,β2β3) and (ββ3, 6β3)
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3. Given π(π₯) = β2π₯ + 3 and π(π₯) = π₯2 β 6π₯ + 3, find the π₯-value(s) that satisfy
π(π₯) = π(π₯).
π(π₯) = π(π₯) means they are equal when the graphs intersect!
β2π₯ + 3 = π₯2 β 6π₯ + 3
0 = π₯2 β 4π₯ Now find π¦-values or π(π₯)!
0 = π₯(π₯ β 4) π¦ = β2(0) + 3 π¦ = β2(4) + 3
π₯ = 0 and π₯ = 4 π¦ = 3 π¦ = β5
Solutions: (0, 3) and (4, β5)
4. Algebraically, determine the points of intersection of (π₯ β 1)2 + (π¦ β 2)2 = 4
and π¦ β 2 = 2π₯.
Substitute: π¦ = 2π₯ + 2
(π₯ β 1)2 + (2π₯ + 2 β 2)2 = 4
(π₯ β 1)(π₯ β 1) + (2π₯)2 = 4 Quadratic Formula?
π₯2 β 2π₯ + 1 + 4π₯2 = 4 Completing the Square?
5π₯2β2π₯ β 3 = 0 AC Method: (5)(β3) = β15
5π₯2β5π₯ + 3π₯ β 3 = 0 β5, 3
5π₯(π₯ β 1) + 3(π₯ β 1) = 0 Grouping
(5π₯ + 3)(π₯ β 1) = 0 Now find π¦-values!
π₯ = β3
5 and 1 π¦ β 2 = 2 (β
3
5) π¦ β 2 = 2(1)
π¦ = 1 π¦ = 4
Solutions: (β3
5, 1) and (1, 4)
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5. Solve the following system of equatons algebraically:
π₯ + 2π¦ = 0
π₯2 β 2π₯ + π¦2 β 2π¦ β 3 = 0
Substitute: π₯ = β2π¦
(β2π¦)2 β 2(β2π¦) + π¦2 β 2π¦ β 3 = 0
4π¦2 + 4π¦ + π¦2 β 2π¦ β 3 = 0 Completing the Square?
5π¦2 + 2π¦ β 3 = 0 AC Method?
π¦ =β(2)Β±β(2)2β4(5)(β3)
2(5) Quadratic Formula: π = 5, π = 2, π = β3
π¦ =β2Β±β4+60
10=β2Β±β64
10 Now find π₯-values!
π¦ =β2Β±8
10 π₯ + 2 (
3
5) = 0 π₯ + 2(β1) = 0
π¦ =3
5 and β 1 π₯ = β
6
5 π₯ = 2
Solutions: (β6
5,3
5) and (2, β1)
Push Yourself! Applications of Systems of Equations DECIMALS use Calculator
6. Sabrina is playing ball with her dog. She throws the ball in a parabolic path
that can be modeled by the function π¦ = β1
2(π₯ β 3)2 + 7. Her brother, Bobby, is
playing in a tree next to her. Bobby shines his laser pointer from the tree in a line
that can be modeled by the function π¦ = β1
2π₯ + 8.5. At what point(s) will the ball
and the laser beam intersect?
β1
2(π₯ β 3)2 + 7 = β
1
2π₯ + 8.5
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Homework Lesson 7.5: Solving Systems of Equations Algebraically
1. Find all solutions of the system of equation algebraically:
(π₯ β 2)2 + (π¦ + 3)2 = 4
π₯ β π¦ = 3
2. Given π(π₯) = β2π₯ + 3 and π(π₯) = π₯2 β 6π₯ + 3, find the π₯-value(s) that satisfy
π(π₯) = π(π₯).
3. Algebraically, determine the points of intersection of βπ¦2 + 6π¦ + π₯ β 9 = 0
and 6π¦ = π₯ + 27
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4. A boy standing on the top of a building in Albany throws a water balloon up
vertically. The height, β (in feet) of the water balloon is given by the equation
β(π‘) = β16π‘2 + 64π‘ + 192, where π‘ is the time (in seconds) after he threw the
water balloon. What is the value of π‘ when the balloon hits the ground? Explain
and show how you arrived at the answer.
5. What is the total number of points of intersection of the graphs of the
equations 2π₯2 β π¦2 = 8 and π¦ = π₯ + 2?
(1) 1 (2) 2 (3) 3 (4) 0
6. Amy solved the equation 2π₯2 + 5π₯ β 42 = 0. She stated that the solutions to
the equation were 7
2 and β6. Do you agree with Amyβs solutions? Explain why or
why not.
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