martin-gay, developmental mathematics 1 warm up factor the following

Martin-Gay, Developmental Mathematics 1

Warm Up

2(3 4)(3 4)x x

218 32x Factor the following

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

Solving Quadratic Equations by the

Quadratic Formula

https://www.youtube.com/watch?v=YCuXiujC3KE


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.


Quadratic Formula

• If we are unable to factor a quadratic function to find the roots we can utilize the Quadratic Formula. The entire equation can tell us the number of roots & the radicand tells us the number of real solutions. A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions


Discriminant

Positive, perfect squares 2 real, rational roots

Positive, not perfect squares 2 real, irrational roots

Zero 1 real rational root

Negative 2 complex roots


Solution


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)

martin-gay, developmental mathematics 1 warm up factor the following

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