name: notes: zero product property & quadratic equations
TRANSCRIPT
1
Name: ___________________________________ NOTES: Zero Product Property & Quadratic Equations
Solving Quadratic Equationsβ¦make sure your equation is set equal to zero!
Example 1: π₯2 + 3π₯ = 0
π₯(π₯ + 3) = 0 factor
π₯ = 0 | π₯ + 3 = 0| set each factor equal to zero and solve
π₯ = {0, β3} list all values of π₯
Example 2: π¦2 = 16
β16 = β16
π¦2 β 16 = 0 set equal to zero first
(π¦ β 4)(π¦ + 4) = 0 factor
π¦ β 4 = 0 | π¦ + 4 = 0 set each factor equal to zero and solve
+4 + 4 | β 4 β 4
π¦ = 4 | π¦ = β4
π¦ = {4, β4} list all value of π¦
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Name: ___________________________________ NOTES contβd: Zero Product Property & Quadratic Equations
Solve:
1. 5π₯ = 0 2. 5(π₯ β 1) = 0
3. (π₯ + 5)(π₯ β 1) = 0 4. (π + 2)(π + 7) = 0
5. π₯3 + 4π₯2 + 3π₯ = 0 6. π2 β 6π + 1 = β4
7. π2 β π β 2 = 0 8. π¦+3
3=
6
π¦
3
Name: ___________________________________ HOMEWORK: Zero Product Property & Quadratic Equations
1. π₯2 β 9 = 0 2. π2 + 16π + 64 = 0
3. π₯3 β π₯2 β 12π₯ = 0 4. π₯+2
2=
12
π₯
5. π2 β 4π = 0 6. π2 β 3π β 4 = 0
7. 4π2 β 12π = β32 + 3π2
4
Name: _____________________________________ CLASSWORK: Quadratics by Factoring
Solve the following by factoring:
1. π₯2 β 2π₯ β 15 = 0 2. 3π₯2 = 48
3. 4π₯2 β 36 = 0 4. π₯2 β 5π₯ = 0
5. π₯2 β 3π₯ = 10 6. π₯2 + 3π₯ β 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. Itβs all about TWO!!!
Step 1: isolate the π₯β²π π₯2 + ππ₯ =?
Step 2: take Β½ of π or Γ· 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 π₯2 β 10π₯ + 16 = 0
Step 1: isolate the π₯β²π π₯2 β 10π₯ = β16
Step 2: take Β½ of π or Γ· 2 1
2ππ β 10 = β5
Step 3: square the result of step 2 (β5)2 = 25
Step 4: add the result of step 3 to both sides π₯2 β 10π₯ + 25 = β16 + 25
Factor: (π₯ β 5)2 = 9
Take the square root of each side: β(π₯ β 5)2 = Β±β9
Solve: π₯ β 5 = Β±3
π₯ = 5 Β± 3
Simplify π₯ = 8 πππ π₯ = 2
2. π₯2 + 12π₯ β 13 = 0 3. π₯2 β 2π₯ β 7 = 0
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Name: _______________________________________ NOTES/HOMEWORK:
Solving Quadratic Equations by Completing the Square
4. 3π§2 β 18π§ β 30 = 0 5. π₯2 β 8π₯ + 7 = 0
What would be the value of π that makes each trinomial a perfect square
6. π2 + 14π + π 7. β2 β 20β + π
8. π2 + 10π + 14 = β7 9. π£2 β 2π£ = 3
7
Name: ____________________________________ CLASSWORK: Complete the Square
Solve by completing the square:
1. π₯2 β 2π₯ β 15 = 0 2. π₯2 + 4π₯ = 96
3. π₯2 + 6π₯ + 9 = 64 4. π₯2 + 10π₯ + 24 = 48
5. (#32 Aug 14.CC) 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.
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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 ππ₯2 + ππ₯ + π = 0 where π β 0 are given by the
formula π₯ =βπΒ±βπ2β4ππ
2π
Always remember to write the equation so that it is in the form of ππ₯2 + ππ₯ + π = 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 βπ2 β 4ππ, the value of π2 β 4ππ must be nonnegative. If π2 β 4ππ is negative, the
equation has no real roots.
Example: Use the quadratic formula to solve π₯2 β 6π₯ β 2 = 0. π = 1, π = β6, and π = β2
π₯ =βπ Β± βπ2 β 4ππ
2π
=β(β6) Β± β(β6)2 β 4(1)(β2)
2(1)
=6 Β± β44
2=
6 Β± 2β11
2= 3 Β± β11 β 6.32 and β 0.32
Solve each equation by using the quadratic formula. Leave it in simplest radical form.
1. π₯2 β 4π₯ + 3 = 0 2. β2π₯2 + 3π₯ + 5 = 0
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Name: _____________________________________ NOTES/HOMEWORK: Quadratic Formula
Solve each equation by using the quadratic formula. Leave it in simplest radical form.
3. 0 = 4π₯ + π₯2 + 2 4. 0 = β3π₯2 + 6π₯ + 9
5. 4π‘2 = 144 6. 2π₯2 β 9π₯ + 4 = 0
7. π2 + 2π β 3 = 0 8. 5π£2 β 21 = 10π£
10
Name: _____________________________________ NOTES/HOMEWORK: Quadratic Formula
Solve each equation by using the quadratic formula. Leave it in simplest radical form.
9. π2 + 2π β 33 = 0 10. 2π2 + 12π + 10 = 0
11. π2 β 12π + 26 = 0 12. 5π2 + 19π β 68 = β2
13. π2 β 8π β 48 = 0 14. 3π₯2 + 20π₯ + 36 = 4
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Name: ___________________________________ CLASSWORK: quadratic formula
Solve each using the quadratic formula:
1. β3π₯2 + π₯ + 14 = 0 2. 2π2 β 98 = 0
3. π₯2 + 7π₯ = β12
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 π₯2 + 4π₯ β 16 = 0?
(1) 2 Β± 2β5 (2) β2 Β± 2β5 (3) 2 Β± 4β5 (4) β2 Β± 4β5
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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, (π2 β 4ππ), is positive, negative, or zero. This expression has
a special name: the discriminant.
If the discriminant is positive β if π2 β 4ππ > 0 β then the quadratic has two solutions.
If the discriminant is zero β if π2 β 4ππ = 0 β then the quadratic equation has one solution.
If the discriminant is negative β if π2 β 4ππ < 0 β then the quadratic equation has no solutions.
Example: How many solutions does the quadratic equation 2π₯2 + 5π₯ + 2 = 0 have?
π = 2, π = 5, and π = 2
π2 β 4ππ = 52 β 4(2)(2) = 25 β 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. π2 + 7π + 33 = 8 β 3π 2. 7π₯2 + 2π₯ + 5 = 0
3. 2π¦2 + 10π¦ = π¦2 + 4π¦ β 3 4. 4π§2 + 9 β 4π§ = 0
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Name: _________________________________ NOTES/HOMEWORK contβd: 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:
(β2)2 β 4(1)(2) (β4)2 β 4(β1)(β4) (β4)2 β 4(1)(0) (β8)2 β 4(β1)(β13)
Graph #: _____ Graph #: _____ Graph #: _____ Graph #: _____
How many solutions does each quadratic equation have? Hint: use π2 β 4ππ
6. βπ2 + 8π = 12 7. 2π₯2 + 4π₯ = 4
8. β3π£2 + 40 = 2π£ 9. 2π2 β π = 36
10. β5π₯2 + 1 = 3π₯ 11. β21π2 β 4 = β10π2 β 12π
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Name: _____________________________ CLASSWORK: Discriminant π2 β 4ππ
1. Find the discriminant and the number of solutions:
(a) π₯2 = 16 (b) 0 = β2π₯2 + 4π₯ β 4 (c) 3π₯2 β 4π₯ = β2
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 + π₯2 = 6π₯
Factor
Quadratic Formula Complete the Square
2. π₯2 + π₯ = 56
Factor
Quadratic Formula Complete the Square
16
Name: _____________________________________ CLASSWORK: Quadratic 3 ways
3. π₯2 + π₯ + 2 = 22
Factor
Quadratic Formula Complete the Square
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Name: _________________________________ UNIT REVIEW
1. Keith determines the zeros of the function π(π₯) to be -6 and 5. What could be Keithβs function?
(1) π(π₯) = (π₯ + 5)(π₯ + 6)
(2) π(π₯) = (π₯ + 5)(π₯ β 6)
(3) π(π₯) = (π₯ β 5)(π₯ + 6)
(4) π(π₯) = (π₯ β 5)(π₯ β 6)
2. Which equation has the same solution as π₯2 β 6π₯ β 12 = 0?
(1) (π₯ + 3)2 = 21
(2) (π₯ β 3)2 = 21
(3) (π₯ + 3)2 = 3
(4) (π₯ β 3)2 = 3
3. What are the roots of the equation π₯2 + 4π₯ β 16 = 0?
(1) 2 Β± 2β5
(2) β2 Β± 2β5
(3) 2 Β± 4β5
(4) β2 Β± 4β5
4. Solve: 7π2 β 14π β 56 = 0 5. Solve: π2 β 9π β 38 = β9
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Name: ____________________________________ Unit Review continued
Find how many solutions there are in questions # 6 & 7:
6. π2 + 13π + 22 = 7 7. π₯2 + 7π₯ β 45 = 7
Solve:
8. 4π₯2 + 16π₯ = 0 9. π₯2 = 3π₯ + 3
10. Write an equation that defines π(π₯) as a trinomial where π(π₯) = (3π₯ β 1)(3 β π₯) + 4π₯2 + 19
Solve for π₯ when π(π₯) = 0.