Chapter 16 Quadratic Equations. Martin-Gay, Developmental Mathematics 2 16.1 – Solving Quadratic Equations by the Square Root Property 16.2 – Solving

Chapter 16

Quadratic Equations

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

§ 16.1

Solving Quadratic Equations by the Square

Root Property


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


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


Example


Solve x2 + 4 = 0 x2 = 4

There is no real solution because the square root of 4 is not a real number.


Example


Solve (x + 2)2 = 25

x = 2 ± 5

x = 2 + 5 or x = 2 – 5

x = 3 or x = 7

5252 x


Example


Solve (3x – 17)2 = 28

72173 x

3

7217 x

7228 3x – 17 =


Example

§ 16.2

Solving Quadratic Equations by Completing

the Square


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


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


Example


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.


Example


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.



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



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



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

§ 16.3

Solving Quadratic Equations by the

Quadratic Formula


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.


A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions.

a

acbbx

2

42



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


Example


)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.


Example


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.


Example


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


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


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.


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


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

§ 16.4

Graphing Quadratic Equations in Two

Variables


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


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)


Example


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


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


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


Example

§ 16.5

Interval Notation, Finding Domain and Ranges from

Graphs, and Graphing Piecewise-Defined Functions


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


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


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


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


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


Domain is {Polar Bear, Cow, Chimpanzee, Giraffe, Gorilla, Kangaroo, Red Fox}

Range is {20, 15, 10, 7}

Domain and Range

Example continued


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.


Example continued

Graphing Piecewise-Defined Functions

x

y

x f (x) = x + 3

1 4

2 5

3 6

x f (x) = 3x – 1

0 – 1(closed circle)

–1 – 4

–2 – 7

(0, –1)

(–1, 4)

(–2, 7)

Open circle (0, 3)

(3, 6)

Documents

Chapter 16 Quadratic Equations. Martin-Gay, Developmental Mathematics 2 16.1 – Solving Quadratic Equations by the Square Root Property 16.2 – Solving