2.1 – Linear and Quadratic Equations Linear Equations.

2.1 – Linear and Quadratic Equations

4456 aa

44306 aa

30430302 a

aaaa 4444306 4302 a

342 a

2

34

2

2

a

17a

Linear Equations


A quadratic equation is written in the Standard Form,

2 0ax bx c where a, b, and c are real numbers and .

Examples:

2 7 12 0x x

23 4 15x x

7 0x x

(standard form)

Quadratic Equations

Zero Factor Property: If a and b are real numbers and if , 0ab

Examples:

7 0x x

then or . 0a 0b

0x 7 0x 7x 0x


Zero Factor Property: If a and b are real numbers and if , 0ab

Examples: 10 3 6 0x x

then or . 0a 0b

10 0x 3 6 0x

10x 3 6x 2x

10 10 01 0x 63 66 0x 3 6

3 3

x


Solving Quadratic Equations: 1) Write the equation in standard form.

4) Solve each equation.

2) Factor the equation completely.

3) Set each factor equal to 0.

5) Check the solutions (in original equation).


2 3 18 0x x

6 0x 3 0x 3x

6x 3x

2 3 18x x

18 :Factors of1,18 2, 9 3, 6

6x 0


23 7 6x x 3 0x 3 2 0x

3 3 2 0x x

3x 2

3x

3 7 6x x

23 7 6 0x x 3 2x

6 :Factors of2, 31, 6

3:Factors of1, 3


23 7 6x x 3 0x 3 2 0x

3 3 2 0x x

3x 2

3x

3 7 6x x

23 7 6 0x x 3 2x


𝑃𝑟𝑜𝑑𝑢𝑐𝑡 :3 ∙6=18𝐹𝑎𝑐𝑡𝑜𝑟𝑠𝑜𝑓 18 :1,18 2,9 3,6𝐹𝑎𝑐𝑡𝑜𝑟𝑠𝑜𝑓 18 h𝑡 𝑎𝑡 𝑐𝑜𝑚𝑏𝑖𝑛𝑒𝑡𝑜7 : 2,93 𝑥2− 2𝑥+9 𝑥−6=0

𝑥 (3 𝑥− 2 )+3 (3 𝑥−2 )=0(3 𝑥− 2 ) (𝑥+3 )=0

3 0x 3 2 0x

3 3 2 0x x

3x 2

3x 3 2x


29 24 16x x 29 24 16 0x x

3 4 0x 3 4 3 4 0x x

4

3x

3 4x

9 16and are perfect squares


32 18 0x x 2x

2 0x

2x

3x 3 0x 3 0x

3x 0x

2 9x 0

3x 3x 0



9 𝑥2− 3𝑥− 12𝑥+4=0

3 𝑥 (3 𝑥− 1 )− 4 (3 𝑥− 1 )¿0

(3 𝑥−1 )(3 𝑥− 4 )¿03 𝑥−1=0 3 𝑥− 4=0

𝑥=13

𝑥=43


3 𝑥3+2𝑥2 −12𝑥− 8=0

𝑥2 (3 𝑥+2 )− 4 (3 𝑥+2 )¿0

(3 𝑥+2 )(𝑥2− 4 )¿0

3 𝑥+2=0

𝑥=−23 𝑥=2

(3 𝑥+2 )(𝑥−2 ) ¿0(𝑥+2 )𝑥−2=0𝑥+2=0

𝑥=−2

23 3 20 7 0x x x

3x

3 0x

7x 7 0x 3 1 0x

1

3x

3x 3 1x

3:Factors of 1, 3 7 :Factors of 1, 7

7x 0 3 1x



3 ( 4 𝑥−11 )=− 12(3 −𝑥)

12𝑥−33 −36+12 𝑥¿−12 𝑥−12 𝑥

𝑓𝑎𝑙𝑠𝑒 𝑠𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡𝑛𝑜𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

−33=− 36

𝐒𝐩𝐞𝐜𝐢𝐚𝐥𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧𝐬


3 (2 𝑥+6 −5 𝑥 )=11𝑥+7 −20 𝑥+11

6 𝑥+18−15 𝑥 − 9𝑥+18¿− 9𝑥+18=− 9𝑥+18

𝑡𝑟𝑢𝑒𝑠𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠

+9𝑥+9 𝑥18=18

𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛

𝐒𝐩𝐞𝐜𝐢𝐚𝐥𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧𝐬

2.2 - Formulas

Example 1:

9, 63d rt t d

63 9r

63 9

9 9

r

7 r

Using the given values, solve for the variable in each formula that was not assigned a value.

7r

Check:

63 9r

63 97

63 63

2.2 - Formulas

Example 2: Volume of a Pyramid1

40, 83

V Bh V h

40 81

3B 1

40 83

3 3 B

120 8B

15 B

120 8

8 8

B

15B

LCD: 3

2.2 - Formulas

Example 3: Solve for the requested variable.

1

22 2A bh

1

2A bh

Area of a Triangle – solve for b

2A bh

2A bh

h h

2Ab

h

LCD: 2

2.2 - FormulasExample 4: Solve for the requested variable.

932

532 32F C 9

325

F C

Celsius to Fahrenheit – solve for C

932

5F C

5 32 9F C

5 32

9

FC

932

55 5F C

5 32 9

9 9

F C

or 532

9F C

LCD: 5


Solve for v

h=𝑣𝑡− 16 𝑡2

+16 𝑡 2+16 𝑡2

h+16 𝑡2=𝑣𝑡

h+16 𝑡 2

𝑡=𝑣

h+16 𝑡 2

𝑡=𝑣𝑡𝑡


Solve for x

𝑎𝑥−5=𝑐𝑥−2−𝑐𝑥−𝑐𝑥𝑎𝑥−𝑐𝑥−5=−2

+5+5𝑎𝑥−𝑐𝑥=3𝑥 (𝑎−𝑐)=3

𝑥=3

𝑎−𝑐

𝑥 (𝑎−𝑐)𝑎−𝑐

= 3𝑎−𝑐

2.3 - ApplicationsSimple Interest

Simple Interest .Principal= Interest Rate

𝑰=𝑷𝑹𝑻

Interest Rate is stated as a percent and converted to a decimal for calculation purposes.

. Time

Time is stated in years or part of a year.

61.25

Simple Interest

7%

Find the simple interest on a five year loan of $875 at a rate of 7%.

.$875

875 . 0.07

$306.25

. 5

. 5

. 5

306.25

𝑰=𝑷𝑹𝑻2.3 - Applications

You invest $8000 in two accounts and earn a total of $323 in interest from both accounts in one year. The interest rates on the accounts were 4.6% and 2.8%. How much was invested in each account?

Total Interest = Account 1 + Account 2

𝑰=𝑷𝑹𝑻

Account 1

Account 2

Principal Interest Rate Interest

x

8000-x

𝟒 .𝟔%=𝟎 .𝟎𝟒𝟔

𝟐 .𝟖%=𝟎 .𝟎𝟐𝟖

𝟎 .𝟎𝟒𝟔𝒙

𝟎 .𝟎𝟐𝟖 (𝟖𝟎𝟎𝟎− 𝒙)

𝟑𝟐𝟑=𝟎 .𝟎𝟒𝟔𝒙+𝟎 .𝟎𝟐𝟖 (𝟖𝟎𝟎𝟎− 𝒙)𝟑𝟐𝟑=𝟎 .𝟎𝟒𝟔𝒙+𝟐𝟐𝟒−𝟎 .𝟎𝟐𝟖𝒙𝟗𝟗=𝟎 .𝟎𝟏𝟖𝒙𝟓𝟓𝟎𝟎=𝒙

Account 1: $5500

Account 2: 8000 – 5500

2.3 - Applications

Account 2: $2500

x + 7

𝒂𝟐+𝒃𝟐=𝒄𝟐x

2.3 - ApplicationsIn a right triangle, the length of the longer leg is 7 more inches than the shorter leg. The length of the hypotenuse is 8 more inches than the length of the shorter leg. Find the length of all three sides.

x + 8

(𝒙)𝟐+(𝒙+𝟕)𝟐=(𝒙+𝟖)𝟐

𝒙𝟐+𝒙𝟐+𝟏𝟒𝒙+𝟒𝟗=𝒙𝟐+𝟏𝟔𝒙+𝟔𝟒−𝒙𝟐−𝟏𝟔𝒙−𝟔𝟒− 𝒙𝟐−𝟏𝟔𝒙−𝟔𝟒𝒙𝟐−𝟐 𝒙−𝟏𝟓=𝟎(𝒙+𝟑)(𝒙−𝟓)¿𝟎𝒙+𝟑=𝟎 𝒙−𝟓=𝟎𝒙=−𝟑𝒙=𝟓

𝒙=𝟓 𝒊𝒏𝒄𝒉𝒆𝒔𝒙+𝟕=𝟏𝟐𝒊𝒏𝒄𝒉𝒆𝒔𝒙+𝟖=𝟏𝟑𝒊𝒏𝒄𝒉𝒆𝒔

𝑺𝒆𝒍𝒍𝒊𝒏𝒈𝒑𝒓𝒊𝒄𝒆×𝒅𝒐𝒘𝒏𝒑𝒂𝒚𝒎𝒆𝒏𝒕𝒑𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆=𝒅𝒐𝒘𝒏𝒑𝒂𝒚𝒎𝒆𝒏𝒕

2.3 - ApplicationsA family paid $26,250 as a down payment for a home. This represents 15% of the selling price. What is the price of the home?

𝒑×𝟎 .𝟏𝟓¿𝟐𝟔𝟐𝟓𝟎𝟎 .𝟏𝟓𝒑=𝟐𝟔𝟐𝟓𝟎𝟎 .𝟏𝟓𝒑𝟎 .𝟏𝟓

=𝟐𝟔𝟐𝟓𝟎𝟎 .𝟏𝟓

𝒑=𝟏𝟕𝟓𝟎𝟎𝟎𝒑=$𝟏𝟕𝟓 ,𝟎𝟎𝟎

2.3 - Applications

𝒙𝟓 𝒙+𝟔

𝒙+𝟓 𝒙+𝟔

Special Pairs of Angles

Complimentary angles: Two angles whose sum is 90°. They are compliments of each other.

Supplementary angles: Two angles whose sum is 180°. They are supplements of each other.

One angle is six more than five times the other angle. What are their measurements if the are supplements of each other?

¿𝟏𝟖𝟎𝟔 𝒙+𝟔=𝟏𝟖𝟎𝟔 𝒙=𝟏𝟕𝟒𝟔𝒙𝟔

=𝟏𝟕𝟒𝟔

𝒙=𝟐𝟗

𝒙=𝟐𝟗°𝟓 (𝟐𝟗)+𝟔=𝟏𝟓𝟏°

𝟐𝟗°+𝟏𝟓𝟏°=𝟏𝟖𝟎°

2.3 - ApplicationsA flower bed is in the shape of a triangle with one side twice the length of the shortest side, and the third side is 30 feet more than the length of the shortest side. Find the dimensions if the perimeter is 102 feet.

x = the length of the shortest side

2x = the length of the second side

x + 30 = the length of the third side

x 2x

x + 30

P = a + b + c

102 = x + 2x + x + 30

102 = 4x + 30

102 – 30 = 4x + 30 – 30

72 = 4x

4

4

4

72 x

𝑥=18 𝑓𝑒𝑒𝑡2(18)=36 𝑓𝑒𝑒𝑡

18+30=48 𝑓𝑒𝑒𝑡

→ 𝑥=18

2.3 - ApplicationsThe length of a rectangle is 4 less than twice the width. The area of the rectangle is 70. Find the dimensions of the rectangle.

h𝑙𝑒𝑛𝑔𝑡 =2 𝑥− 4

𝐴=𝑙×𝑤70=(2 𝑥− 4 ) 𝑥70=2 𝑥2 − 4 𝑥0=2 𝑥2− 4 𝑥− 700=2 (𝑥2 −2 𝑥−35)0=2 (𝑥+5)(𝑥− 7)

(𝑥+5 )=0 (𝑥−7 )=0𝑥=−5 𝑥=7

h𝑙𝑒𝑛𝑔𝑡 =2 𝑥− 4=2 (7 )− 4=10

2.3 - Applications

35 mph

𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆=𝒓𝒂𝒕𝒆× 𝒕𝒊𝒎𝒆40 mph

Two cars leave an airport at the same time. One is traveling due north at a rate of 40 miles per hour and the other is travelling due east at a rate of 35 miles per hour. When will the distance between the two cars be 110 miles?

110 mi.𝒅𝒏=𝟒𝟎𝒕𝒂𝟐+𝒃𝟐=𝒄𝟐

(𝟒𝟎𝒕 )𝟐+(𝟑𝟓𝒕 )𝟐=𝟏𝟏𝟎𝟐

𝟐𝟖𝟐𝟓𝒕𝟐

𝟐𝟖𝟐𝟓=𝟏𝟐𝟏𝟎𝟎

𝟐𝟖𝟐𝟓

𝒅𝒆=𝟑𝟓𝒕

𝟏𝟔𝟎𝟎 𝒕𝟐+𝟏𝟐𝟐𝟓𝒕𝟐=𝟏𝟐𝟏𝟎𝟎𝟐𝟖𝟐𝟓 𝒕𝟐=𝟏𝟐𝟏𝟎𝟎

𝒕𝟐=𝟒 .𝟐𝟖𝟑 𝒕=𝟐 .𝟎𝟕𝒉𝒓𝒔 .→

It takes Karen 3 hours to row a boat 30 kilometers upstream in a river. If the current was 4 kilometers per hour, how fast would she row in still water?

Rate Equation:

Rate upstream: (𝑑=𝑟𝑡 ) 30=𝑟𝑢 (3) 𝑟𝑢=10 h𝑘𝑝

Rate in still water: 𝑟 𝑠=10 h𝑘𝑝 +4 h𝑘𝑝 𝑟 𝑠=14 h𝑘𝑝

How long would it take her to row 30 kilometers in still water? (𝑑=𝑟𝑡 ) 30=14 𝑡 𝑡=2.14 h𝑟𝑠 .How long would it take her to row 30 kilometers downstream? (𝑑=𝑟𝑡 ) 30=(14+4)𝑡 𝑡=1.67h𝑟𝑠 .30=18 𝑡

2.3 - Applications

2.4 – Linear Inequalities in One Variable

An inequality is a statement that contains one of the symbols: < , >, ≤ or ≥.

Equations Inequalities

x = 3

12 = 7 – 3y

x > 3

12 ≤ 7 – 3y

A solution of an inequality is a value of the variable that makes the inequality a true statement.

The solution set of an inequality is the set of all solutions.



Example

Graph each set on a number line and then write it in interval notation.

a. { | 2}

b. { | 1}

c. { | 0.5 3}

x x

x x

x x

a. [2, )

b.

c. (0.5,3]


Addition Property of Inequality

If a, b, and c are real numbers, then

a < b and a + c < b + c

a > b and a + c > b + c

are equivalent inequalities.

Also,

If a, b, and c are real numbers, then

a < b and a - c < b - c

a > b and a - c > b - c



ExampleSolve: Graph the solution set.3 4 2 6x x

{ | 10} or 10,x x

[


Multiplication Property of Inequality

If a, b, and c are real numbers, and c is positive, thena < b and ac < bc are equivalent inequalities.

If a, b, and c are real numbers, and c is negative, thena < b and ac > bc



The direction of the inequality sign must change when multiplying or dividing by a negative value.

Solve: Graph the solution set.

{ | 3} or 3,x x

2.3 6.9x

The inequality symbol is reversed since we divided by a negative number.

(

Example


Solve: 3x + 9 ≥ 5(x – 1). Graph the solution set.

3x + 9 ≥ 5x – 53x – 3x + 9 ≥ 5x – 3x – 5

9 ≥ 2x – 5

14 ≥ 2x7 ≥ x

9 + 5 ≥ 2x – 5 + 5

3x + 9 ≥ 5(x – 1)

x ≤ 7

[


ExampleSolve: 7(x – 2) + x > –4(5 – x) – 12. Graph the solution set.

7(x – 2) + x > –4(5 – x) – 12

7x – 14 + x > –20 + 4x – 12

8x – 14 > 4x – 32

8x – 4x – 14 > 4x – 4x – 32

4x – 14 > –32

4x – 14 + 14 > –32 + 14

4x > –18

x > –4.5

(


Intersection of Sets

The solution set of a compound inequality formed with and is the intersection of the individual solution sets.


Compound Inequalities

Example

Find the intersection of: {2,4,6,8} {3,4,5,6}

The numbers 4 and 6 are in both sets.

The intersection is {4, 6}.



Solve and graph the solution for x + 4 > 0 and 4x > 0.

Example

First, solve each inequality separately.

x + 4 > 0

x > – 4

4x > 0

x > 0and

-4(

0(

( (0, )



Example

0 4(5 – x) < 8

0 20 – 4x < 8

0 – 20 20 – 20 – 4x < 8 – 20

– 20 – 4x < – 12

5 x > 3

Remember that the sign direction changes when you divide by a number < 0!

(

[

(3,5]


3 4 5


Example – Alternate Method

0 4(5 – x)

0 20 – 4x

0 – 20 20 – 20 – 4x

– 20 – 4x

5 x (

[

(3,5]


3 4 5

4(5 – x) < 8

20 – 4x < 8

20 – 20 – 4x < 8 – 20

– 4x < – 12

x > 3

0 4(5 – x) < 8

Dividing by negative:

change sign

Dividing by negative:

change sign


The solution set of a compound inequality formed with or is the union of the individual solution sets.

Union of Sets



Find the union of:

Example

{2,4,6,8} {3,4,5,6}

The numbers that are in either set are {2, 3, 4, 5, 6, 8}.

This set is the union.



Example: Solve and graph the solution for 5(x – 1) –5 or 5 – x < 11

5(x – 1) –5

5x – 5 –5

5x 0x 0

5 – x < 11

–x < 6

x > – 6

or

0[

-6(

(–6, )-6(



Example:

or

,



2.1 – Linear and Quadratic Equations Linear Equations.

Documents

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applications simple

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