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. 2 −4=0 Graphically Factoring Square Root Method DOPS ( − 2)( + 2) = 0 − 2 = 0 + 2 = 0 = ±2 2 =4 √ 2 = √4 = ±2 2) Express in simplest radical form: √−25 √−1 ∗ √25 5 √−32 √−16 ∗ √2 4√2 √ 4 5 2 ∗ √5 √5 ∗ √5 2√5 5 Advice to solving Quadratic Equations: Express quadratic in standard form 2 + + = 0 Determine appropriate methods to solve quadratic equations.
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Lesson 7.1: Solving Quadratic Equations by Graphing ......1 Lesson 7.1: Solving Quadratic Equations by Graphing, Factoring, and Square Root Method Learning Goals: 1) How do we solve
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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. 𝑥2 − 4 = 0
Graphically Factoring Square Root Method
DOPS (𝑥 − 2)(𝑥 + 2) = 0 𝑥 − 2 = 0 𝑥 + 2 = 0
𝑥 = ±2
𝑥2 = 4
√𝑥2 = √4 𝑥 = ±2
2) Express in simplest radical form:
√−25
√−1 ∗ √25 5𝑖
√−32
√−16 ∗ √2
4𝑖√2
√4
5
2 ∗ √5
√5 ∗ √5
2√5
5
Advice to solving Quadratic Equations:
Express quadratic in standard form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
Determine appropriate methods to solve quadratic equations.
2
1. Solve the following equation by factoring: 𝑥2 − 48 = 2𝑥
Set = 0
𝑥2 − 2𝑥 − 48 = 0
(𝑥 − 8)(𝑥 + 6) = 0
𝑥 − 8 = 0 𝑥 + 6 = 0
𝑥 = 8 & − 6
2. Solve for 𝑥 and express in simplest form: 3𝑥2 + 9 = 0
3𝑥2 = −9
𝑥2 = −3
𝑥 = ±𝑖√3
3. Use graphing to find the roots of 𝑥2 − 2𝑥 = 8
1. Solve for 𝑎: 𝑎2 = 5𝑎
𝑎2 − 5𝑎 = 0 set = 0
𝑎(𝑎 − 5) = 0 GCF = 𝑎
𝑎 = 0 𝑎 − 5 = 0
𝑎 = 0 & 5
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2. Find the zeros of 𝑓(𝑥) = 5𝑥2 − 8𝑥 − 4
0 = 5𝑥2 − 8𝑥 − 4 AC method
0 = 5𝑥2 − 10𝑥 + 2𝑥 − 4 5(−4) = −20
0 = 5𝑥(𝑥 − 2) + 2(𝑥 − 2) −10, 2
0 = (5𝑥 + 2)(𝑥 − 2)
𝑥 = −2
5 & 2
3. Solve for 𝑥: (𝑥 + 3)2 = 16
√(𝑥 + 3)2 = √16
𝑥 + 3 = ±4
𝑥 = −3 ± 4
−3 + 4 − 3 − 4
𝑥 = 1 & − 7
4. Algebraically solve for 𝑥 in simplest form: 3𝑥2 + 4 = 0
3𝑥2 = −4
𝑥2 = −4
3
𝑥 = ±√−4
√3= ±
2𝑖
√3 ∗√3
√3 or ±
2𝑖√3
3
5. Find the roots of the equation: 2𝑥2 + 𝑥 − 3 = (𝑥 − 1)(𝑥 + 2)
2𝑥2 + 𝑥 − 3 = 𝑥2 + 2𝑥 − 𝑥 − 2
2𝑥2 + 𝑥 − 3 = 𝑥2 + 𝑥 − 2
𝑥2 − 1 = 0
𝑥 = ±1
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6. Find the zeros of 𝑥 − 4 =5
𝑥 Cross Multiply!
𝑥(𝑥 − 4) = 5
𝑥2 − 4𝑥 = 5
𝑥2 − 4𝑥 − 5 = 0
(𝑥 − 5)(𝑥 + 1) = 0
𝑥 = 5 & − 1
7. Solve for 𝑥: 1
4(𝑥 − 6)2 = 8
(𝑥 − 6)2 = 32
𝑥 − 6 = ±√32
𝑥 = 6 ± 4√2
8. Solve for 𝑥: (𝑥 + 1)2 − 87 = 𝑥2
(𝑥 + 1)(𝑥 + 1) − 87 = 𝑥2
𝑥2 + 2𝑥 + 1 − 87 = 𝑥2
2𝑥 − 86 = 0
2𝑥 = 86
𝑥 = 43
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Check for understanding…
1. On a test, Samantha says that the roots of the equation 𝑥2 + 2𝑥 − 3 = 0 are
−1 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.
(−1)2 + 2(−1) − 3 = 0 (3)2 + 2(3) − 3 = 0
1 − 2 − 3 = 0 9 + 6 − 3 = 0
−4 ≠ 0 12 ≠ 0
Therefore she is not correct!
(𝑥 + 3)(𝑥 − 1) = 0
𝑥 = −3 & 1
2. Phil’s teacher gave the class the quadratic function 𝑓(𝑥) = (𝑥 − 2)2 − 4.
a) State two different methods Phil could use to solve the equation 𝑓(𝑥) = 0.
b) Using one of the methods stated in part a, solve 𝑓(𝑥) = 0 for 𝑥.
Factoring, Quadratic Formula
𝑓(𝑥) = (𝑥 − 2)(𝑥 − 2) − 4
𝑓(𝑥) = 𝑥2 − 4𝑥 + 4 − 4
𝑓(𝑥) = 𝑥(𝑥 − 4)
𝑥 = 0 & 4
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Homework Lesson 7.1: Solving Quadratic Equations by Graphing,
Factoring, and Square Root Method
1. Solve for 𝑥: 𝑥 − 4 =−3
𝑥
2. Find the roots of 3𝑥2 − 5𝑥 = 36 − 2𝑥
3. Solve for 𝑥 in simplest form: 3𝑥2 + 25 = 0
4. Solve for 𝑥: 5(𝑥 − 1)2 = 50
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5. Use graphing to solve the given
equation for 𝑥: 𝑥2 + 5𝑥 + 6 = 0
6. Find the zeros of 𝑓(𝑥) = 4𝑥2 − 16 − 20
7. Solve for 𝑥 in simplest form: (2𝑥 + 3)(𝑥 − 4) = 2𝑥2 + 13𝑥 + 15
8. Solve for 𝑥: 𝑥2 + 𝑥 − 1 = (−4𝑥 + 3)2
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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 𝑐 that makes the expression a perfect square
trinomial. Then write the expression as the square of a binomial.
𝑥2 + 14𝑥 + 𝑐
𝑐 = (14
2)2
= 72 = 49
𝑥2 + 14𝑥 + 49 (𝑥 + 7)(𝑥 + 7) (𝑥 + 7)2
𝑥2 − 9𝑥 + 𝑐
𝑐 = (−9
2)2
=81
4
𝑥2 − 9𝑥 +81
4
(𝑥 −9
2)2
Steps to Completing the Square
1. The "𝑎" coefficient must equal 1 (divide all terms by "𝑎").
2. Isolate the 𝑥-terms (move "𝑐" to the other side).
3. Take (𝑏
2)2 and add that number to each side.
4. Factor the perfect square trinomial and express it as (𝑥 + 𝑎)2.
5. Solve for 𝑥 by using the square root method.
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Completing the Square when 𝑎 = 1
Example 1: Solve 𝑥2 − 10𝑥 + 13 = 0 by completing the square.
Solution Steps
𝑥2 − 10𝑥 + 13 = 0 Write original equation.
𝑥2 − 10𝑥 = −13 Write left side in the form 𝑥2 + 𝑏𝑥, I say “isolate the 𝑥”
𝑥2 − 10𝑥 + 𝑐 = −13 + 𝑐
𝑐 = (−10
2)
2
= (−5)2 = 25
Complete the square, I say “divide by two and square it”
𝑥2 − 10𝑥 + 25 = −13 + 25
(𝑥 − 5)2 = 12
Write left side as a binomial squared.
√(𝑥 − 5)2 = √12 Take the square roots of each side.
𝑥 − 5 = ±√12
𝑥 = 5 ± √12
Solve for 𝑥.
𝑥 = 5 ± 2√3 Simplify
Practice: Solve 𝑥(𝑥 + 3) = −2 by completing the square, and express the results
in simplest form.
𝑥2 + 3𝑥 = −2
𝑥2 + 3𝑥 + 𝑐 = −2 + 𝑐
𝑐 = (3
2)2=9
4
𝑥2 + 3𝑥 +9
4= −2 +
9
4
(𝑥 +3
2)2=1
4
𝑥 +3
2= ±
1
2
𝑥 = −3
2±1
2
𝑥 = −1 & − 2
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Completing the Square when 𝑎 ≠ 1
Example 1: Solve 3𝑥2 − 12𝑥 + 27 = 0 by completing the square.
Solution Steps
3𝑥2 − 12𝑥 + 27 = 0 Write original equation.
𝑥2 − 4𝑥 + 9 = 0 Get 𝑎 = 1before you can complete the square! Divide each side by the coefficient
of 𝑥2. 𝑥2 − 4𝑥 = −9 Write left side in the form 𝑥2 + 𝑏𝑥.
𝑥2 − 4𝑥 + 𝑐 = −9 + 𝑐
𝑐 = (−4
2)2
= (−2)2 = 4
Complete the square.
𝑥2 − 4𝑥 + 4 = −9 + 4
(𝑥 − 2)2 = −5
Write left side as a binomial squared.
√(𝑥 − 2)2 = √−5 Take the square roots of each side.
𝑥 − 2 = ±√−5
𝑥 = 2 ± √−5
Solve for 𝑥.
𝑥 = 2 ± 𝑖√5 Simplify
Practice: Solve 4𝑥2 − 20𝑥 = −9 by completing the square, and express the
results in simplest form.
𝑥2 − 5𝑥 = −9
4 Isolate the 𝑥 and make sure 𝑎 = 1
𝑥2 − 5𝑥 + 𝑐 = −9
4+ 𝑐
𝑐 = (−5
2)2=25
4
𝑥2 − 5𝑥 +25
4= −
9
4+25
4
(𝑥 −5
2)2= 4
𝑥 −5
2= ±2
𝑥 =5
2± 2
𝑥 =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) 𝑥2 − 9𝑥 − 1 = 0 b) 5𝑥(𝑥 + 6) = −50
𝑥2 − 9𝑥 = 1 5𝑥2 + 30𝑥 = −50
𝑥2 − 9𝑥 + 𝑐 = 1 + 𝑐 𝑥2 + 6𝑥 = −10
𝑐 = (−9
4)2=81
16 𝑥2 + 6𝑥 + 𝑐 = −10 + 𝑐
𝑥2 − 9𝑥 +81
16= 1 +
81
16 𝑐 = (
6
2)2= (3)2 = 9
(𝑥 −9
4)2=97
16 𝑥2 + 6𝑥 + 9 = −10 + 9
𝑥 −9
4= ±
√97
4 (𝑥 + 3)2 = −1
𝑥 =9
4±√97
4 𝑥 + 3 = ±𝑖
𝑥 = −3 ± 𝑖
Analysis Question: Which equation has the same solutions as 𝑥2 + 6𝑥 − 7 = 0?