Name: NOTES: Zero Product Property & Quadratic Equations

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Name: ___________________________________ NOTES: Zero Product Property & Quadratic Equations

Solving Quadratic Equations…make sure your equation is set equal to zero!

Example 1: 𝑥2 + 3𝑥 = 0

𝑥(𝑥 + 3) = 0 factor

𝑥 = 0 | 𝑥 + 3 = 0| set each factor equal to zero and solve

𝑥 = {0, −3} list all values of 𝑥

Example 2: 𝑦2 = 16

−16 = −16

𝑦2 − 16 = 0 set equal to zero first

(𝑦 − 4)(𝑦 + 4) = 0 factor

𝑦 − 4 = 0 | 𝑦 + 4 = 0 set each factor equal to zero and solve

+4 + 4 | − 4 − 4

𝑦 = 4 | 𝑦 = −4

𝑦 = {4, −4} list all value of 𝑦

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Name: ___________________________________ NOTES cont’d: Zero Product Property & Quadratic Equations

Solve:

1. 5𝑥 = 0 2. 5(𝑥 − 1) = 0

3. (𝑥 + 5)(𝑥 − 1) = 0 4. (𝑔 + 2)(𝑔 + 7) = 0

5. 𝑥3 + 4𝑥2 + 3𝑥 = 0 6. 𝑘2 − 6𝑘 + 1 = −4

7. 𝑘2 − 𝑘 − 2 = 0 8. 𝑦+3

3=

6

𝑦

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Name: ___________________________________ HOMEWORK: Zero Product Property & Quadratic Equations

1. 𝑥2 − 9 = 0 2. 𝑔2 + 16𝑔 + 64 = 0

3. 𝑥3 − 𝑥2 − 12𝑥 = 0 4. 𝑥+2

2=

12

𝑥

5. 𝑛2 − 4𝑛 = 0 6. 𝑎2 − 3𝑎 − 4 = 0

7. 4𝑝2 − 12𝑝 = −32 + 3𝑝2

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Name: _____________________________________ CLASSWORK: Quadratics by Factoring

Solve the following by factoring:

1. 𝑥2 − 2𝑥 − 15 = 0 2. 3𝑥2 = 48

3. 4𝑥2 − 36 = 0 4. 𝑥2 − 5𝑥 = 0

5. 𝑥2 − 3𝑥 = 10 6. 𝑥2 + 3𝑥 − 40 = 0

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Name: _______________________________________ NOTES:

Solving Quadratic Equations by Completing the Square

Completing the Square is used to make any quadratic a perfect square. It’s all about TWO!!!

Step 1: isolate the 𝑥′𝑠 𝑥2 + 𝑏𝑥 =?

Step 2: take ½ of 𝑏 or ÷ 2

Step 3: square the result of step 2

Step 4: add the result of step 3 to both sides

Example 1. This is not a perfect square 𝑥2 − 10𝑥 + 16 = 0

Step 1: isolate the 𝑥′𝑠 𝑥2 − 10𝑥 = −16

Step 2: take ½ of 𝑏 or ÷ 2 1

2𝑜𝑓 − 10 = −5

Step 3: square the result of step 2 (−5)2 = 25

Step 4: add the result of step 3 to both sides 𝑥2 − 10𝑥 + 25 = −16 + 25

Factor: (𝑥 − 5)2 = 9

Take the square root of each side: √(𝑥 − 5)2 = ±√9

Solve: 𝑥 − 5 = ±3

𝑥 = 5 ± 3

Simplify 𝑥 = 8 𝑎𝑛𝑑 𝑥 = 2

2. 𝑥2 + 12𝑥 − 13 = 0 3. 𝑥2 − 2𝑥 − 7 = 0

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Name: _______________________________________ NOTES/HOMEWORK:

Solving Quadratic Equations by Completing the Square

4. 3𝑧2 − 18𝑧 − 30 = 0 5. 𝑥2 − 8𝑥 + 7 = 0

What would be the value of 𝑐 that makes each trinomial a perfect square

6. 𝑞2 + 14𝑞 + 𝑐 7. ℎ2 − 20ℎ + 𝑐

8. 𝑚2 + 10𝑚 + 14 = −7 9. 𝑣2 − 2𝑣 = 3

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Name: ____________________________________ CLASSWORK: Complete the Square

Solve by completing the square:

1. 𝑥2 − 2𝑥 − 15 = 0 2. 𝑥2 + 4𝑥 = 96

3. 𝑥2 + 6𝑥 + 9 = 64 4. 𝑥2 + 10𝑥 + 24 = 48

5. (#32 Aug 14.CC) A student was given the equation 𝑥2 + 6𝑥 − 13 = 0 to solve by completing the square. The

first step that was written is shown below.

𝑥2 + 6𝑥 = 13

The next step in the student’s process was 𝑥2 + 6𝑥 + 𝑐 = 13 + 𝑐.

State the value of 𝑐 that creates a perfect square trinomial.

Explain how the value of 𝑐 is determined.

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Name: _____________________________________ NOTES/HOMEWORK: Quadratic Formula

The method of completing the square can be used to develop a general formula called the quadratic

formula that can be used to sole any quadratic equation.

The Quadratic Formula

The roots of a quadratic equation of the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 where 𝑎 ≠ 0 are given by the

formula 𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎

Always remember to write the equation so that it is in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0!!

Parentheses are important to use when substituting the values for the variables in the quadratic equation.

In order to find a real value for √𝑏2 − 4𝑎𝑐, the value of 𝑏2 − 4𝑎𝑐 must be nonnegative. If 𝑏2 − 4𝑎𝑐 is negative, the

equation has no real roots.

Example: Use the quadratic formula to solve 𝑥2 − 6𝑥 − 2 = 0. 𝑎 = 1, 𝑏 = −6, and 𝑐 = −2

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

=−(−6) ± √(−6)2 − 4(1)(−2)

2(1)

=6 ± √44

2=

6 ± 2√11

2= 3 ± √11 ≈ 6.32 and − 0.32

Solve each equation by using the quadratic formula. Leave it in simplest radical form.

1. 𝑥2 − 4𝑥 + 3 = 0 2. −2𝑥2 + 3𝑥 + 5 = 0

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3. 0 = 4𝑥 + 𝑥2 + 2 4. 0 = −3𝑥2 + 6𝑥 + 9

5. 4𝑡2 = 144 6. 2𝑥2 − 9𝑥 + 4 = 0

7. 𝑎2 + 2𝑎 − 3 = 0 8. 5𝑣2 − 21 = 10𝑣

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9. 𝑟2 + 2𝑟 − 33 = 0 10. 2𝑛2 + 12𝑛 + 10 = 0

11. 𝑚2 − 12𝑚 + 26 = 0 12. 5𝑛2 + 19𝑛 − 68 = −2

13. 𝑘2 − 8𝑘 − 48 = 0 14. 3𝑥2 + 20𝑥 + 36 = 4

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Name: ___________________________________ CLASSWORK: quadratic formula

Solve each using the quadratic formula:

1. −3𝑥2 + 𝑥 + 14 = 0 2. 2𝑛2 − 98 = 0

3. 𝑥2 + 7𝑥 = −12

4. (#11 June 13) The solutions of 28162 xx are

(1) -2 and -14 (2) 2 and 14 (3) -4 and -7 (4) 4 and 7

5. (#10 June 14.CC) What are the roots of the equation 𝑥2 + 4𝑥 − 16 = 0?

(1) 2 ± 2√5 (2) −2 ± 2√5 (3) 2 ± 4√5 (4) −2 ± 4√5

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Name: __________________________________ NOTES: Quadratic Formula & The Discriminant

As we have seen, there can be 0, 1, or 2 solutions to a quadratic equation, depending on whether the

expression inside the square root sign, (𝑏2 − 4𝑎𝑐), is positive, negative, or zero. This expression has

a special name: the discriminant.

If the discriminant is positive – if 𝑏2 − 4𝑎𝑐 > 0 – then the quadratic has two solutions.

If the discriminant is zero – if 𝑏2 − 4𝑎𝑐 = 0 – then the quadratic equation has one solution.

If the discriminant is negative – if 𝑏2 − 4𝑎𝑐 < 0 – then the quadratic equation has no solutions.

Example: How many solutions does the quadratic equation 2𝑥2 + 5𝑥 + 2 = 0 have?

𝑎 = 2, 𝑏 = 5, and 𝑐 = 2

𝑏2 − 4𝑎𝑐 = 52 − 4(2)(2) = 25 − 16 = 9 > 0

Thus, the quadratic equation has 2 solutions.

For Exercises #1 - 4, without solving, determine the number of real solutions for each quadratic

equation. Then, state the roots for #3!

1. 𝑝2 + 7𝑝 + 33 = 8 − 3𝑝 2. 7𝑥2 + 2𝑥 + 5 = 0

3. 2𝑦2 + 10𝑦 = 𝑦2 + 4𝑦 − 3 4. 4𝑧2 + 9 − 4𝑧 = 0

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Name: _________________________________ NOTES/HOMEWORK cont’d: Discriminant

5. State whether the discriminant of each quadratic equation is positive, negative, or equal to zero

on the line below the graph. Then identify which graph matches the discriminants below:

___________ __________ __________ __________

Discriminant A: Discriminant B: Discriminant C: Discriminant D:

(−2)2 − 4(1)(2) (−4)2 − 4(−1)(−4) (−4)2 − 4(1)(0) (−8)2 − 4(−1)(−13)

Graph #: _____ Graph #: _____ Graph #: _____ Graph #: _____

How many solutions does each quadratic equation have? Hint: use 𝑏2 − 4𝑎𝑐

6. −𝑏2 + 8𝑏 = 12 7. 2𝑥2 + 4𝑥 = 4

8. −3𝑣2 + 40 = 2𝑣 9. 2𝑘2 − 𝑘 = 36

10. −5𝑥2 + 1 = 3𝑥 11. −21𝑎2 − 4 = −10𝑎2 − 12𝑎

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Name: _____________________________ CLASSWORK: Discriminant 𝑏2 − 4𝑎𝑐

1. Find the discriminant and the number of solutions:

(a) 𝑥2 = 16 (b) 0 = −2𝑥2 + 4𝑥 − 4 (c) 3𝑥2 − 4𝑥 = −2

2. How many roots exist for the following equation: 42 2 xx

(a) 0 (b) 1 (c) 2 (d) 3

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Name: _____________________________________ CLASSWORK: Quadratic 3 ways

Solve each equation 3 ways: Factor, Quadratic Formula, and Complete the Square

(HINT: you should get the same answer 3 times)

1. 8 + 𝑥2 = 6𝑥

Factor

Quadratic Formula Complete the Square

2. 𝑥2 + 𝑥 = 56

Factor


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Name: _____________________________________ CLASSWORK: Quadratic 3 ways

3. 𝑥2 + 𝑥 + 2 = 22

Factor


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Name: _________________________________ UNIT REVIEW

1. Keith determines the zeros of the function 𝑓(𝑥) to be -6 and 5. What could be Keith’s function?

(1) 𝑓(𝑥) = (𝑥 + 5)(𝑥 + 6)

(2) 𝑓(𝑥) = (𝑥 + 5)(𝑥 − 6)

(3) 𝑓(𝑥) = (𝑥 − 5)(𝑥 + 6)

(4) 𝑓(𝑥) = (𝑥 − 5)(𝑥 − 6)

2. Which equation has the same solution as 𝑥2 − 6𝑥 − 12 = 0?

(1) (𝑥 + 3)2 = 21

(2) (𝑥 − 3)2 = 21

(3) (𝑥 + 3)2 = 3

(4) (𝑥 − 3)2 = 3

3. What are the roots of the equation 𝑥2 + 4𝑥 − 16 = 0?

(1) 2 ± 2√5

(2) −2 ± 2√5

(3) 2 ± 4√5

(4) −2 ± 4√5

4. Solve: 7𝑏2 − 14𝑏 − 56 = 0 5. Solve: 𝑟2 − 9𝑟 − 38 = −9

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Name: ____________________________________ Unit Review continued

Find how many solutions there are in questions # 6 & 7:

6. 𝑛2 + 13𝑛 + 22 = 7 7. 𝑥2 + 7𝑥 − 45 = 7

Solve:

8. 4𝑥2 + 16𝑥 = 0 9. 𝑥2 = 3𝑥 + 3

10. Write an equation that defines 𝑚(𝑥) as a trinomial where 𝑚(𝑥) = (3𝑥 − 1)(3 − 𝑥) + 4𝑥2 + 19

Solve for 𝑥 when 𝑚(𝑥) = 0.

Name: NOTES: Zero Product Property & Quadratic Equations

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