Section 10.1 Solving Quadratic Equations by the Square Root Property

Section 10.1

Solving Quadratic Equations by the Square Root Property

OBJECTIVESSolve quadratic equations of the form

B2

± = AX B C

A X 2 = A

DEFINITION

Square Root Property of Equations

X = A or X = – A

If A is a positive number andX 2 = A, then X = ± A:

PROCEDURETo solve any equation of form

AX 2 – B = 0

AX 2 =B

Add B

Then

PROCEDURE

X 2 = B

A

Divide by A

To solve any equation of form

AX 2 – B = 0

PROCEDURE

Using square root property,

X = BA

, BA0

To solve any equation of form

AX 2 – B = 0

Section 10.1Exercise #1

Chapter 10Quadratic Equations

Solve.

x 2 = 64

x = ± 64

x = ± 8



Solve.

49x 2 – 25 = 0

49x 2 = 25

x 2 = 25

49

x = ± 5

7



Solve.

6x 2 + 49 = 0

6x 2 = – 49

x 2 = –

49

6

x = –

49

6

No real-number solution.



Solve. 236 + 1 – 7 = 0x

236 + 1 = 7x

2 7 + 1 = 36

x

x + 1 = ± 7

36

x + 1 = ±

7

6

Solve. 236 + 1 – 7 = 0x

x = – 1 ±

7

6or

x =

– 6 ± 7

6

x = – 1 ±

7

6 =

– 6 ± 7

6



Solve.

29 – 3 + 1 = 0x

29 – 3 = – 1x

2 1– 3 = 9 – x

x – 3 = –

1

9

No real-number solution.

Section 10.2

Solving Quadratic Equations by Completing the Square

OBJECTIVES

A Solve a quadratic equation by completing the square.

PROCEDURE

1. Find the coefficient of x term.

b

Completing the Square x2 + bx +

PROCEDURE

2. Divide coefficient by 2.

2

b


PROCEDURE

3. Square this number to obtain last term.

2 2 =

2 4

b b


PROCEDURE

1. Write equation with variables in descending order on left and constants on right.

Solving a Quadratic Equation by Completing the Square

PROCEDURESolving a Quadratic Equation by Completing the Square

2. If coefficient of square term is not 1, divide each term by this coefficient.


3. Add square of one-half of

coefficient of first-degree term to both sides.


4. Rewrite left-hand side as a

perfect square binomial.


5. Use square root property to solve resulting equation.

Section 10.2




2 2

Find the missing term in the expr

+ 4 = x + 8x +

ession

.x

22•

1= 8 = 4 = 16

2



22 – 8x +

The missing terms in the expressions

are ___ and _______ respectiv

.

=

ely

x

2

2• –

1 = 8 = = 16

2 ( ) – 4

16 4x

2 28 16 ( 4) x x x



To solve the equation 8x2 – 32x = – 5 by completing thesquare, the first step will be to divide eachterm by______. 8



To solve the equation x2 – 8x = – 15 by completing thesquare, one has to add ______ to both sides ofthe equation.

2

2•

1 – 8 = – 4 = 16

2

16

Section 10.3

Solving Quadratic Equations by the Quadratic Formula

OBJECTIVESA Write a quadratic

equation in the form

ax2 + bx + c = 0 andidentify a, b, and c.

OBJECTIVESB Solve a quadratic

equation using the quadratic formula.

DEFINITIONThe Quadratic Formula

2The solutions of

+ + = 0 (a 0) areax bx c

– b± b2 – 4ac2a

DEFINITION

– b + b2 – 4ac2a

and

– b – b2 – 4ac2a

The Quadratic Formula

DEFINITION

2If – 4 0, the equationhas real-number solutions.

b ac

If b2 – 4ac<0, the equationhas no real-number solutions.


DEFINITION

Start by writing equation instandard form.


2If – 4 0 is a perfectsquare, try factoring.

b ac

(Note: Some perfect squares are 0, 1, 4, 9, 16 and so on)



The solution of ax 2 + bx + c = 0 is .

2 – ± – 4 = , 0

2b b ac

x aa



2x2 – 3x – 2 = 0

x =

– b ± b2 – 4ac2a

Solve.

a = 2, b = – 3, c = – 2

2 – – 3 ± – 3 – 4 2 – 2=

2 2

2x2 – 3x – 2 = 0

Solve.

a = 2, b = – 3, c = – 2

=

3 ± 9 + 164

=

3 ± 54

2 – – 3 ± – 3 – 4 2 – 2=

2 2

2x2 – 3x – 2 = 0

Solve.

a = 2, b = – 3, c = – 2

x = 2 or –

12

=

84

or – 24

=

3 ± 54



x2 = 3x – 2

x2 – 3x + 2 = 0

Solve.

– 2 – 1 = 0x x

x – 2 = 0 x – 1 = 0or

x = 2 or x = 1

Solutions: x = 1, 2



16x = x2

0 = x2 – 16x

Solve.

x = 0 x – 16 = 0or

x = 0 or x = 16

Solutions: x = 0, 16

0 = – 16x x



x2

2 +

54

x = – 12

2x2 + 5x = – 2

Solve.

2x + 1 = 0 x + 2 = 0or

2x = – 1 or x = – 2

Solutions: x = –

12

, – 2

2x2 + 5x + 2 = 0

2 + 1 + 2 = 0x x

LCD = 4

Multiply by 4:

Section 10.4

Graphing Quadratic Equations

OBJECTIVESA Graph quadratic

equations.

OBJECTIVESB Find intercepts and

vertex and graph parabolas involving factorable quadratic expressions.

DEFINITIONGraph of a Quadratic Equation

1. Opens upward if a> 0

2. Opens downward if a < 0

The graph of y = ax2 + bx + c is a parabola that

PROCEDUREGraphing a Factorable Quadratic Equation

1. Find y-intercept by letting x = 0, then finding y.


2. Find x-intercept by letting y = 0, factoring equation then solving for x.


3. Find vertex by averaging solutions of equation and

substituting in equation to find y-coordinate.


4. Plot points found and one or two more points.

Curve drawn through points found is graph.



y

x

5

5 – 5

– 5

Graph y = 3x2 .

0, 0V turns up

when x = ± 1, y = 3

1, 3

– 1, 3 y = 3x2



y

x

5

5 – 5

– 5

2Graph = – 1 + 1.y x

1, 1V turns up

when x = 0, y = 2

0, 2

2, 2 y = x – 1 2

+ 1so

2 = – + y a x h k



y

x

5

5 – 5

– 5

2Graph the equation = – – 1 + 1.y x

1, 1V turns down

when x = 0, y = 0

0, 0

2, 0 2 = – 1 + 1 – y x

so

2 = – + y a x h k



Graph the equation y = – x2 – 2x + 8. Label the vertexand intercepts.

at x intercepts y = 0

0 = – x2 – 2x + 8

0 = + 4 – 2x x

0 = x2 + 2x – 8

x + 4 = 0 x – 2 = 0or

x = – 4 x = 2

Graph the equation y = – x2 – 2x + 8. Label the vertexand intercepts.

at Vertex: x =

– 4 + 22

= – 22

= – 1

Substitute – 1 for x in original equation

= – 1 + 2 + 8

2 = – – 1 – 2 – 1 + 8y

= 9

Vertex – 1, 9

at y intercept x = 0, y = 8

y

x

10

10 – 10

– 10

Vertex – 1, 9

– 1, 9

Intercepts – 4, 0 , 2, 0 , 0, 8

– 4, 0 2, 0

0, 8

Section 10.5

Applications: Pythagoras’ Theorem

OBJECTIVES

A Use the Pythagorean Theorem to solve right triangles.

OBJECTIVES

B Solve word problems involving quadratic equations.

DEFINITION

Pythagorean Theorem

h2 = a2 + b2

Square of hypotenuse of a right triangle equals sum of squares of other two sides:



Find the length of the hypotenuse of a right triangle if the lengths of the two sides are 2 inches and 5 inches.

2 2 2 + = 0a b h h

a = 2, b = 5

h2 = 22 + 52

= 4 + 25

= 29

h = 29 inches



The formula d = 5t2 + v0t gives the distance d (in meters)an object thrown downward with an initial velocityv0 will have gone after t seconds. How longwould it take an object dropped (v0 = 0)from a distance of 180 meters to hitthe ground?

20 = 5 + 0d t v t t

If dropped v0 = 0, so d = 5t2

180 = 5t2

It takes 6 seconds.

when d = 180,

36 = t2

6 = t

Section 10.6

Functions

OBJECTIVES

A Find the domain and range of a relation.

OBJECTIVES

B Determine whether a given relation is a function.

OBJECTIVES

C Use function notation.

D Solve an application.

DEFINITIONS

Relation, Domain and Range

A relation is a set ofordered pairs.

DEFINITIONS


Domain of a relation is the set of all possible x-values.

DEFINITIONS


Range of relation is the set of all possible y-values.

DEFINITIONFunction

Set of pairs in which each domain value has exactly one range value.

(no two different ordered pairs have same first coordinate)



Find the domain and range of

– 1, 1 , – 2, 1 , – 3, 2

Domain = – 1, – 2, – 3

Range = 1, 2



Find the domain and range of

x , y y =

1x – 8

Domain = Real numbers except for 8,

x ° 8

Range = Real numbers except for 0,

y ° 0



State whether each of the following is a function.

a. 2, – 4 , – 1, 3 , – 1, 4

b. 2, – 4 , – 1, 3 , 1, – 3

This is not a function.

When = – 1, there are 2 values for x y

This is a function.

(For every x, unique y)



3 2– 2 = – 2 + 2 – 2 – – 2 – 1f

3 2If = + 2 – – 1, find – 2 .f x x x x f

3 2Given = + 2 – – 1f x x x x

= – 8 + 2 4 + 2 – 1

= – 8 + 8 + 2 – 1

= 1



= 25 – 0.4 dollarsP n n

Given: = 25 – 0.4P n n

The average price P(n) of books depends on the number n of millions of books sold and is given by the function

Find the average price of a book when 20 million copies are sold.

20 = 25 – 0.4 20P

= 25 – 8 = 17

The average price is $17.

Documents

Section 10.1 Solving Quadratic Equations by the Square Root Property