Section 4.3 Solving Quadratics by Factoring and the Square ...
Transcript of Section 4.3 Solving Quadratics by Factoring and the Square ...
Pre-Calculus 11 Updated January 2020
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Section 4.3 β Solving Quadratics by Factoring and the Square Root Method
This Booklet Belongs to: Block:
Definition of a Quadratic Equation An equation that can be written in the form:
πππ + ππ + π = π Where π, π, πππ π are Real Numbers with π β 0
The solutions of the quadratic equation π(π₯) = ππ₯2 + ππ₯ + π = 0 are called zeros, roots, or solutions
The points where the graph crosses the π β ππππ are real solutions, because π(π) is π
The π β ππππππππππ are called the real roots of the quadratic function
A quadratic function can cross the π₯ β ππ₯ππ either π, π, ππ π πππππ
Real Zeros of a Quadratic Function If π(π₯) is a quadratic function and π is a real zero of π(π₯), then the following statements are equivalent
1. π₯ = π ππ π ππππ ππ π‘βπ ππ’πππ‘πππ π(π₯) 2. π₯ = π ππ π ππππ ππ π‘βπ ππ’πππ‘πππ π(π₯)
3. π₯ = π ππ π ππππππππ ππ π‘βπ πππ’ππ‘πππ π(π₯) = 0
4. (π₯ β π) ππ π ππππ‘ππ ππ π‘βπ ππ’πππππ‘ππ πππ’ππ‘πππ π(π₯)
5. (π, 0)ππ ππ π₯ β πππ‘ππππππ‘ ππ π‘πβ ππππβ ππ π(π₯)
Example 1:
π(π₯) = π₯2 β 2π₯ + 3 π(π₯) = π₯2 β 2π₯ + 1 π(π₯) = π₯2 β 2π₯ β 3
No zeros
(No real roots)
One zero
(Double root/two roots equal)
Two zeros
(Two unequal real roots)
Pre-Calculus 11 Updated January 2020
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Factoring Quadratics in the Form ππ + ππ + π
Consider this: (π₯ + π)(π₯ + π) = π₯2 + ππ₯ + ππ₯ + ππ
π₯2 + (π + π)π₯ + ππ
By looking at this we see that:
The first term is the product of π₯ and π₯
The coefficient of the middle term is the sum of π and π
The last term is the product of π and π
This leads us to the general rule:
When factoring π₯2 + ππ₯ + π, look for two factors of c, that multiply to the coefficient of the last
term, and add to the coefficient of the middle term.
Example: Factor π₯2 + 7π₯ + 12
Solution: What two numbers add to 7 and multiply to 12?
Integers that multiply to 12: (1, 12) (2, 6) (3, 4) (β1, β12) (β2, β6) (β3, β4)
Only integers 3 πππ 4 πππ π‘π 7
Therefore π₯2 + 7π₯ + 12 = (π₯ + 3)(π₯ + 4)
We can check our answer using FOIL: (π₯ + 3)(π₯ + 4)
= π₯2 + 3π₯ + 4π₯ + 12
= π2 + ππ + ππ
Example: Factor π₯2 + 8 β 6π₯
Solution: First arrange the polynomial in descending order of powers
π₯2 + 8 β 6π₯ = π₯2 β 6π₯ + 8
β4 and β2 add to β6 and multiply to +8
Therefore: π₯2 β 6π₯ + 8 = (π₯ β 4)(π₯ β 2)
We can check using FOIL
Example: Factor 5π₯2 + 35π₯ + 60
Solution: Always look for a common factor first. The largest common factor is 5
Therefore: 5π₯2 + 35 + 60 = 5(π₯2 + 7 + 12)
= 5(π₯ + 3)(π₯ + 4)
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Example: Factor β π₯2 + 5π₯ + 6
Solution: First factor out β1, so that the coefficient of π₯2 becomes +1.
So β π₯2 + 5π₯ + 6 becomes β (π₯2 β 5π₯ β 6), now factor (π₯2 β 5π₯ β 6)
β6 and 1 multiply to β6 and add to β5
Therefore β π₯2 + 5π₯ + 6 = β(π₯2 β 5π₯ β 6) = β(π₯ β 6)(π₯ + 1)
Note the factors are: (π β π)(π + π) and βπ
SUMMARY OF FACTORING POLYNOMIALS
1. Arrange the polynomial in descending order of powers
2. When the last term is positive, the factors of c are both positive, or both negative. If the
middle term is positive, both integers are positive. If the middle term is negative, both
integers are negative.
Example: π₯2 + 7π₯ + 12 = (π₯ + 4)(π₯ + 3)
The last term is positive, and the middle term is positive, therefore
the factors of 12 are both positive.
Opposite if the middle term was negative and the last positive.
3. When the last term is negative, the factors of c have opposite signs. The larger numeric value
takes the sign of the coefficient of the middle term.
Example: π₯2 β π₯ β 6 = (π₯ β 3)(π₯ + 2)
The last term is negative, therefore the signs of the factor of 6 are
opposite of each other, and since the middle term is negative the
larger numeric value has a negative sign.
Example: π₯2 + 2π₯ β 15 = (π₯ + 5)(π₯ β 3)
The last term is negative, therefore the signs of the factor of 15 are
opposite of each other, and since the middle term is positive the
larger numeric value has a positive sign.
Pre-Calculus 11 Updated January 2020
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Factoring Quadratics in the Form πππ + ππ + π
There are a number of ways to factor equations in this form
It is my humble opinion that the one shown below works the most efficiently
In the last section we factored ππ₯2 + ππ₯ + π where π = π. In this section π will have an
integer value greater than 1.
The βACβ Method of Factoring
The βACβ method is a technique used to factor trinomials
A trinomial consists of three terms (ππ₯Β² + ππ₯ + π).
π π π
1. 6π₯Β² + 7π₯ + 2
1. Multiply the coefficient of the π term with the coefficient of the π term and rewrite the
polynomial with the starting term now π₯2
So 6π₯Β² + 7π₯ + 2 becomes π₯Β² + 7π₯ + 12
3.
a) Find two numbers that multiply to +12 and add to +7
These are: + 3 πππ + 4
b) Rewrite the factored form of the new version of the Polynomial
(π₯ + 3)(π₯ + 4)
c) Now divide the two factors by the original π term and simplify the fractions
(π₯ +3
6)(π₯ +
4
6) β (π₯ +
1
2)(π₯ +
2
3)
d) If you are not left with a denominator then you are done. If a denominator
remains, rewrite it in front of the π₯ term.
(π₯ +1
2)(π₯ +
2
3)
4. Rewrite as the factored from: (2π₯ + 1)(3π₯ + 2)
This is the Factored Form.
Canβt simplify
so write the
2 in front of
the x
Canβt simplify
so write the
3 in front of
the x
Pre-Calculus 11 Updated January 2020
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Special Factors and Zero Product
Factoring Perfect Square Trinomials
0 = π2 + 2ππ + π2 = (π + π)2 Example: π₯2 + 8π₯ + 16 = (π₯ + 4)2
0 = π2 β 2ππ + π2 = (π β π)2 Example: π₯2 β 8π₯ + 16 = (π₯ β 4)2
Factoring a Difference of Squares
0 = π2 β π2 = (π + π)(π β π) Example: π₯2 β 4 = (π₯ β 2)(π₯ + 2) = 0
Example: 16π₯2 β 25π¦2 = (4π₯ + 5π¦)(4π₯ β 5π¦) = 0
The Zero Product
0 = π2 + ππ = π(π + π) Example: 2π₯2 + 6π₯ = 0 β 2π₯(π₯ + 3) = 0
So: π₯ = 0 ππ π₯ = β3
Solving Using Factoring
Example: Solve the equation π₯2 β π₯ β 6 = 0
Solution:
Example: Solve the equation 3π₯2 + 9π₯ = 0
Solution:
Example: Solve the equation 6π₯2 β 7π₯ β 5 = 0
Solution:
(π₯ β 3)(π₯ + 2) = 0
(π₯ β 3) = 0 ππ (π₯ + 2) = 0
π₯ = 3 ππ π₯ = β2
3π₯2 + 9π₯ = 0 β 3π₯(π₯ + 3) = 0
3π₯ = 0 ππ (π₯ + 3) = 0
π₯ = 0 ππ π₯ = β3
6π₯2 β 7π₯ β 5 = 0 β π₯2 β 7π₯ β 30 = 0
(π₯ β 10)(π₯ + 3) = 0 β (π₯ β10
6) (π₯ +
3
6) = 0
(π₯ β5
3) (π₯ +
1
2) = 0 β π₯ =
5
3 ππ π₯ = β
1
2
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Example: Solve the equation π₯(3π₯ + 1) = 2
Solution:
Example: π₯
π₯β5β
3
π₯+1=
30
π₯2β4π₯β5
Solution:
π₯(3π₯ + 1) = 2 β 3π₯2 + π₯ = 2
3π₯2 + π₯ β 2 = 0 β π₯2 + π₯ β 6 = 0 β (π₯ + 3)(π₯ β 2)
(π₯ +3
3) (π₯ β
2
3) = 0 β π₯ = β
3
3= β1 ππ π₯ =
2
3
π₯
π₯ β 5β
3
π₯ + 1=
30
π₯2 β 4π₯ β 5
π₯
π₯ β 5β
3
π₯ + 1=
30
(π₯ β 5)(π₯ + 1)
(π₯ β 5)(π₯ + 1) [π₯
π₯ β 5β
3
π₯ + 1=
30
(π₯ β 5)(π₯ + 1)]
(π₯ + 1)π₯ β 3(π₯ β 5) = 30
π₯2 + π₯ β 3π₯ + 15 β 30 = 0
π₯2 β 2π₯ β 15 = 0 β (π₯ β 5)(π₯ + 3)
π = π ππ π = βπ
When finished inset your answers into the original equation to make sure they are valid
The denominator cannot equal zero when you plug it in
So it this case reject π = π
Factor the denominators
Identify LCM
In this case: (π₯ β 5)(π₯ + 1)
Multiply each term by the
LCM
Reduce
Simplify
Factor
Solve
Pre-Calculus 11 Updated January 2020
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Solving Quadratics Using the Square Root Method
The factor method is definitely the most efficient method of solving quadratics
But not all quadratics can be factored easily of at all
What we get are two methods depending on the situation
The Square Root Method
The Quadratic Equation
The square root method in mainly used when π = 0 in the equation ππ₯2 + ππ₯ + π = 0
Isolate the π₯2 on the left side and square root the other side
Example:
π₯2 β 16 = 0 π₯2 β 16 = 0
(π₯ + 4)(π₯ β 4) or π₯2 = 16
π₯ β 4 = 0 ππ π₯ + 4 = 0 βπ₯2 = Β±β16
π₯ = 4 ππ π₯ = β4 π₯ = Β±4
The procedure on the right is the SQUARE ROOT METHOD
Solving a Quadratic Equation of the form πππ + π = π Step 1: Isolate the ππ on the left side of the equation and the constant on the right Step 2: Take the square root of both sides, the square root of the constant has to be Β± Step 3: Simplify if possible Step 4: Check the solution in the original equation
The Square Root Property is defined as follows:
The Square Root Property The equation π₯2 = π has exactly 2 real solutions
π₯ = βπ πππ π₯ = ββπ ππ π > 0. πβπ π πππ’π‘πππ πππ π€πππ‘π‘ππ π₯ = Β±π
π₯ = 0 ππ π = 0
π₯ = β ππ π < 0 (π€π πππβ²π‘ π‘πππ π‘βπ π ππ’πππ ππππ‘ ππ π πππππ‘ππ£π)
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Example: Solve 4π₯2 β 9 = 0
Solution:
Example: Solve 2π₯2 + 7 = 0
Solution:
Example: Solve 4π₯2 β 7 = 0
Solution:
Example: Solve (π₯ + 2)2 = 10
Solution:
4π₯2 = 9 - Add nine to both sides
π₯2 =9
4 - Divide both sides by 4
π₯ = Β±β9
4= Β±
3
2 - Square Root both sides (remember Β±)
Solutions are: π
ππππ β
π
π
2π₯2 = β7 - Subtract seven from both sides
π₯2 = β7
2 - Divide both sides by 2
π₯ = Β±ββ7
2= β - β denotes the βπππππ πππβ
- We canβt square root a negative
4π₯2 = 7 - Add seven to both sides
π₯2 =7
4 - Divide both sides by 4
π₯ = Β±β7
4= Β±
β7
2 - Square Root both sides, and simplify
Solutions are:
βπ
ππππ β
βπ
π
(π₯ + 2) = Β±β10 - Square Root both sides first, the βsquared termβ is isolated to start
π₯ = β2 Β±β10 - Subtract both sides by 2
Solutions are: βπ + βππ πππ β π β βππ
Solutions are: π΅πππ
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Example: Solve 9(π₯ β 1)2 = 13
Solution:
Example: Solve 4(π₯ + 3)2 + 11 = 0
Solution:
(π₯ β 1)2 =13
9 - Divide both sides by 9 to isolate the βsquared termβ
π₯ β 1 = Β±β13
9 - Square Root both sides
π₯ = 1 Β± β13
9 - Add 1 to both sides
π₯ = 1 Β±β13
3 - Simplify
Solutions are: π +βππ
π πππ π β
βππ
π
(π₯ + 3)2 = β11
4 - Subtract 11 and Divide both sides by 4 to isolate the βsquared termβ
π₯ + 3 = Β±ββ11
4 - Square Root both sides
π₯ = β - Canβt Square Root a Negative
Solutions are: π΅πππ
Pre-Calculus 11 Updated January 2020
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Section 4.3 β Practice Problems
Find the π₯ β πππ‘ππππππ‘π by factoring the following equations.
1. π¦ = π₯2 β 3π₯ β 4
2. π¦ = π₯2 + π₯ β 6
3. π¦ = βπ₯2 + 4
4. π¦ = β1
2π₯2 β π₯ + 4
5. π¦ = 2π₯2 + 5π₯ β 3
6. π¦ = β1
3π₯2 + 3
Solve the following equations. Check your solutions to make sure they are correct
7. 2π₯(4π₯ β 3) = 0
8. (0.25π¦ β 2)(0.2π¦ + 1) = 0
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9. π₯2 = βπ₯
10. π₯2 + 1 = 0
11. 4π¦2 = π¦
12. 3π₯2 β π₯ = 0
13. π₯2 + 5π₯ + 6 = 0
14. π₯2 β 4π₯ + 3 = 0
15. π¦2 + π¦ β 12 = 0
16. π§3 β 16π§ = 0
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17. π§(π§ β 5) = β4
18. (π₯ β 12)(π₯ + 1) = β40
19. (π¦ β 6)(π¦ + 1) = β10
20. π§3 β π§2 = 6π§
21. π₯3 β 3π₯ = 2π₯2
22. π₯
18
2+
π₯
6= 1
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23. (2π₯ β 1)2 = 16
24. (3π₯ + 8)(π₯ β 1) = (π₯ β 1)(π₯ + 3)
25. (2π¦)2 + (π¦ + 5)2 = (2π¦ + 4)2
26. 6π₯2(3π₯ β 1) β π₯(3π₯ β 1) = 2(3π₯ β 1)
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27. 1
π₯+
3
π₯ β 2=
5
8
28. 4
5 +π¦ =
4π¦ β 50
5π¦ β 25
29. 4
π₯2 β 4 β
1
π₯ β 2 = 1
30. 1
π₯ β 3 β
12
π₯2 β 9 = 1
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Complete the square and solve using the square root method, give exact answers
31. π₯2 + 6π₯ = β5
32. π¦2 β 5π¦ + 3 = 0
33. π§2 β 8π§ + 3 = 0
34. π₯2 β 2π₯ β 1 = 0
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35. π¦2 + 4π¦ + 2 = 0
36. π§2 + 2π§ + 7 = 0
37. 5π₯2 β 3π₯ = 1
38. 3π₯2 β π₯ = 3
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39. 3π₯2 = β8π₯ β 2
40. 3π₯2 = 6π₯ + 2
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Section 4.3 β Answer Key
1. (4, 0) πππ (β1, 0)
2. (β3, 0) πππ (2, 0)
3. (β2, 0) πππ (2, 0)
4. (β4, 0) πππ (2, 0)
5. (β3, 0) πππ (1
2, 0)
6. (β3, 0) πππ (3, 0)
7. π₯ = 0 πππ π₯ =3
4
8. π¦ = 8 πππ π¦ = β5
9. π₯ = 0 πππ π₯ = β1
10. ππ ππππ’π‘πππ
11. π¦ = 0 πππ π¦ =1
4
12. π₯ = 0 πππ π₯ =1
3
13. π₯ = β3 πππ π₯ = β2
14. π₯ = 1 πππ π₯ = 3
15. π¦ = β4 πππ π¦ = 3
16. π§ = 0 πππ π§ = 4 πππ π§ = β4
17. π§ = 4 πππ π§ = 1
18. π₯ = 4 πππ π₯ = 7
19. π¦ = 1 πππ π¦ = 4
20. π§ = 0, π§ = 3, πππ π§ = β2
21. π₯ = 0, π₯ = 3, πππ π₯ = β1
22. π₯ = β6 πππ π₯ = 3
23. π₯ =5
2 πππ π₯ = β
3
2
24. π₯ = β5
2 πππ π₯ = 1
25. π¦ = 3
26. π₯ =2
3 πππ π₯ = β
1
2 πππ π₯ =
1
3
27. π₯ = 8 πππ π₯ =2
5
28. π¦ = 2 πππ π¦ = 3
29. π₯ = β3
30. π₯ = 0 πππ π₯ = 1
31. π₯ = β1 πππ π₯ = β5
32. π¦ =5+β13
2 and π¦ =
5ββ13
2
33. π§ = 4 + β13 and π§ = 4 β β13
34. π₯ = 1 + β2 and π₯ = 1 β β2
35. π¦ = β2 + β2 and π¦ = β2 β β2
36. ππ ππππ’π‘πππ
37. π₯ =3+β29
10 and π₯ =
3ββ29
10
38. π₯ =1+β37
6 and π₯ =
1ββ37
6
39. π₯ =β4+β10
3 and π₯ =
β4ββ10
3
40. π₯ = 1 + β5
3 and π₯ = 1 β β
5
3