Post on 21-Dec-2015
Martin-Gay, Developmental Mathematics 2
16.1 – Solving Quadratic Equations by the Square Root Property
16.2 – Solving Quadratic Equations by Completing the Square
16.3 – Solving Quadratic Equations by the Quadratic Formula
16.4 – Graphing Quadratic Equations in Two Variables
16.5 – Interval Notation, Finding Domains and Ranges from Graphs and Graphing Piecewise-Defined Functions
Chapter Sections
Martin-Gay, Developmental Mathematics 4
Square Root Property
We previously have used factoring to solve quadratic equations.
This chapter will introduce additional methods for solving quadratic equations.
Square Root PropertyIf b is a real number and a2 = b, then
ba
Martin-Gay, Developmental Mathematics 5
Solve x2 = 49
2x
Solve (y – 3)2 = 4
Solve 2x2 = 4
x2 = 2
749 x
y = 3 2
y = 1 or 5
243 y
Square Root Property
Example
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Solve x2 + 4 = 0 x2 = 4
There is no real solution because the square root of 4 is not a real number.
Square Root Property
Example
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Solve (x + 2)2 = 25
x = 2 ± 5
x = 2 + 5 or x = 2 – 5
x = 3 or x = 7
5252 x
Square Root Property
Example
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Solve (3x – 17)2 = 28
72173 x
3
7217 x
7228 3x – 17 =
Square Root Property
Example
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In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left.
Also, the constant on the left is the square of the constant on the right.
So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples).
Completing the Square
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What constant term should be added to the following expressions to create a perfect square trinomial?
x2 – 10xadd 52 = 25
x2 + 16xadd 82 = 64
x2 – 7x
add 4
49
2
72
Completing the Square
Example
Martin-Gay, Developmental Mathematics 12
We now look at a method for solving quadratics that involves a technique called completing the square.
It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section.
Completing the Square
Example
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Solving a Quadratic Equation by Completing a Square
1) If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient.
2) Isolate all variable terms on one side of the equation.
3) Complete the square (half the coefficient of the x term squared, added to both sides of the equation).
4) Factor the resulting trinomial.
5) Use the square root property.
Completing the Square
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Solve by completing the square.
y2 + 6y = 8y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
y = 3 ± 1
y = 4 or 2
y + 3 = ± = ± 11
Solving Equations
Example
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Solve by completing the square.
y2 + y – 7 = 0
y2 + y = 7
y2 + y + ¼ = 7 + ¼
2
29
4
29
2
1y
2
291
2
29
2
1 y
(y + ½)2 = 429
Solving Equations
Example
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Solve by completing the square.
2x2 + 14x – 1 = 0
2x2 + 14x = 1
x2 + 7x = ½
2
51
4
51
2
7x
2
517
2
51
2
7 x
x2 + 7x + = ½ + = 4
49
4
49
4
51
(x + )2 = 4
51
2
7
Solving Equations
Example
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The Quadratic Formula
Another technique for solving quadratic equations is to use the quadratic formula.
The formula is derived from completing the square of a general quadratic equation.
Martin-Gay, Developmental Mathematics 19
A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions.
a
acbbx
2
42
The Quadratic Formula
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Solve 11n2 – 9n = 1 by the quadratic formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1
)11(2
)1)(11(4)9(9 2
n
22
44819
22
1259
22
559
The Quadratic Formula
Example
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)1(2
)20)(1(4)8(8 2
x
2
80648
2
1448
2
128 20 4 or , 10 or 22 2
x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = 20
8
1
2
5Solve x2 + x – = 0 by the quadratic formula.
The Quadratic Formula
Example
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Solve x(x + 6) = 30 by the quadratic formula.
x2 + 6x + 30 = 0
a = 1, b = 6, c = 30
)1(2
)30)(1(4)6(6 2
x
2
120366
2
846
So there is no real solution.
The Quadratic Formula
Example
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The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant.
The discriminant will take on a value that is positive, 0, or negative.
The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.
The Discriminant
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Use the discriminant to determine the number and type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
There are no real solutions.
The Discriminant
Example
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Solving Quadratic Equations
Steps in Solving Quadratic Equations1) If the equation is in the form (ax+b)2 = c, use
the square root property to solve.
2) If not solved in step 1, write the equation in standard form.
3) Try to solve by factoring.
4) If you haven’t solved it yet, use the quadratic formula.
Martin-Gay, Developmental Mathematics 26
Solve 12x = 4x2 + 4.
0 = 4x2 – 12x + 4
0 = 4(x2 – 3x + 1)
Let a = 1, b = -3, c = 1
)1(2
)1)(1(4)3(3 2
x
2
493
2
53
Solving Equations
Example
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Solve the following quadratic equation.
02
1
8
5 2 mm
0485 2 mm
0)2)(25( mm
02025 mm or
25
2 mm or
Solving Equations
Example
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We spent a lot of time graphing linear equations in chapter 3.
The graph of a quadratic equation is a parabola.
The highest point or lowest point on the parabola is the vertex.
Axis of symmetry is the line that runs through the vertex and through the middle of the parabola.
Graphs of Quadratic Equations
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x
y
Graph y = 2x2 – 4.
x y
0 –4
1 –2
–1 –2
2 4
–2 4
(2, 4)(–2, 4)
(1, –2)(–1, – 2)
(0, –4)
Graphs of Quadratic Equations
Example
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Although we can simply plot points, it is helpful to know some information about the parabola we will be graphing prior to finding individual points.
To find x-intercepts of the parabola, let y = 0 and solve for x.
To find y-intercepts of the parabola, let x = 0 and solve for y.
Intercepts of the Parabola
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If the quadratic equation is written in standard form, y = ax2 + bx + c,
1) the parabola opens up when a > 0 and opens down when a < 0.
2) the x-coordinate of the vertex is . a
b
2
To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y.
Characteristics of the Parabola
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x
yGraph y = –2x2 + 4x + 5.
x y
1 7
2 5
0 5
3 –1
–1 –1
(3, –1)(–1, –1)
(2, 5)(0, 5)
(1, 7)Since a = –2 and b = 4, the graph opens down and the x-coordinate of the vertex is 1
)2(2
4
Graphs of Quadratic Equations
Example
§ 16.5
Interval Notation, Finding Domain and Ranges from
Graphs, and Graphing Piecewise-Defined Functions
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Recall that a set of ordered pairs is also called a relation.
The domain is the set of x-coordinates of the ordered pairs.
The range is the set of y-coordinates of the ordered pairs.
Domain and Range
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Find the domain and range of the relation {(4,9), (–4,9), (2,3), (10, –5)}
• Domain is the set of all x-values, {4, –4, 2, 10}• Range is the set of all y-values, {9, 3, –5}
Example
Domain and Range
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Find the domain and range of the function graphed to the right. Use interval notation. x
y
Domain is [–3, 4]
Domain
Range is [–4, 2]
Range
Example
Domain and Range
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Find the domain and range of the function graphed to the right. Use interval notation. x
y
Domain is (– , )
DomainRange is [– 2, )
Range
Example
Domain and Range
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Input (Animal)• Polar Bear• Cow• Chimpanzee• Giraffe• Gorilla• Kangaroo• Red Fox
Output (Life Span)
20
15
10
7
Find the domain and range of the following relation.
Example
Domain and Range
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Domain is {Polar Bear, Cow, Chimpanzee, Giraffe, Gorilla, Kangaroo, Red Fox}
Range is {20, 15, 10, 7}
Domain and Range
Example continued
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Graph each “piece” separately.
Graph3 2 if 0
( ) . 3 if 0
x xf x
x x
Graphing Piecewise-Defined Functions
Example
Continued.
x f (x) = 3x – 1
0 – 1(closed circle)
–1 – 4
–2 – 7
x f (x) = x + 3
1 4
2 5
3 6
Values 0. Values > 0.