special equations : and / or and quadratic inequalities and / or are logic operators. and – where...

Special Equations : AND / OR and Quadratic Inequalities

AND / OR are logic operators.

AND – where two solution sets “share” common elements.

- similar to intersection of two sets

OR – where two solution sets are merged together

- similar to union of two sets







When utilizing these in graphing multiple inequality equations, a number line graph helps to “see” the final solution.







When utilizing these in graphing multiple inequality equations, a number line graph helps to “see” the final solution.

We will first look at how they are different.


EXAMPLE : 5 AND 3for set solution theShow xx



STEPS : 1. Create a number line and locate your points.

( open circle for 5 and closed for – 3 )

- when graphing, graph one point higher than the other

- 3 5






2. Graph the solution set for each

- 3 5







3. Find where they “share” elements

- where are they “on top” of each other

- in this case they share between (– 3) and 5

- 3 5










4. This shared space is our final graph

- 3 5











- 3 5

Answer as an interval


EXAMPLE # 2 : 5 AND 3for set solution theShow xx


5 AND 3for set solution theShow xx




- 3 5

EXAMPLE # 2 :







- 3 5

EXAMPLE # 2 :






2. Graph the solution set for each 3. Find where they “share” elements - where are they “on top” of each other - in this case they share numbers greater than 5

- 3 5

EXAMPLE # 2 :







- 3 5



EXAMPLE # 3: 5 AND 3for set solution theShow xx




- 3 5







- 3 5

2. Graph the solution set for each3. Find where they “share” elements - where are they “on top” of each other - in this case they DO NOT share elements






- 3 5


3. Find where they “share” elements - where are they “on top” of each other - in this case they DO NOT share elements

4. SO we have An EMPTY SET

Ø


EXAMPLE # 4 : 5 OR 3for set solution theShow xx






- 3 5






2. Graph the solution for each

- 3 5







3. Now merge the two graphs and keep everything

- this will be your answer

- 3 5









- 3 5



Graphing Quadratic Inequalities :

1. Factor to find “critical points” ( where the equation = 0 )

2. Locate your points on a number line

3. Pick a test point for TRUE or FALSE

- true / false changes every time you pass a critical point







Example # 1 : Graph the solution set for 062 xx








2 3

02 03 023

xx

xxxx








2 3

02 03 023

xx

xxxx

- 2 3

Open Circle








2 3

02 03 023

xx

xxxx

- 2 3

Open Circle

TRUE 06

0600

0 : TEST2

x

TTTTTTTTTTTTTTT FFFFFFFFFFFF








2 3

02 03 023

xx

xxxx

- 2 3

Open Circle

TRUE 06

0600

0 : TEST2

x

TTTTTTTTTTTTTTT FFFFFFFFFFFFALWAYS graph the TRUE sections…









4,3 043 xxxx








4,3 043 xxxx

- 4 - 3

Closed Circle








4,3 043 xxxx

- 4 - 3

Closed Circle

TRUE 012

012070

0Test 2

x

FFFFF TTTTTTTTTTTTTTTTTTTTTT

ALWAYS graph the TRUE sections…


In some cases, the variable will be located in the denominator of a fraction. Since the denominator can not equal zero, zero instantly becomes a critical point with an open circle above it…

Example # 1 : Graph the solution set for 318

x

x

0



Next, multiply EVERYTHING by “x” and get all terms on one side…

Example # 1 : Graph the solution set for

0183

318

318

2

2

xx

xx

xxx

xx

0




Now just follow the steps as we did before…


0183

318

318

2

2

xx

xx

xxx

xx

0

3,6

036

xx

xx

-3 6

Closed Circle





** I like to test 1000 in these…because 18/1000 is very small, you can disregard it !!!


0183

318

318

2

2

xx

xx

xxx

xx

0

3,6

036

xx

xx

-3 6

Closed Circle

1Test x

FALSE 31000

31000

181000

1000Test

x

FTFT







0183

318

318

2

2

xx

xx

xxx

xx

0

3,6

036

xx

xx

-3 6

Closed Circle

1Test x

FALSE 31000

31000

181000

1000Test

x

FTFTALWAYS graph the TRUE sections…




0

815

x

x





0

0158

815

815

2

2

xx

xx

xxx

xx




Factor and get critical points…


0

0158

815

815

2

2

xx

xx

xxx

xx

3,5

035

xx

xx

3 5

Open Circle





Test 1000…


0

0158

815

815

2

2

xx

xx

xxx

xx

3,5

035

xx

xx

3 5

Open Circle

FALSE 81000

81000

151000

FT TF





Test 1000…


0

0158

815

815

2

2

xx

xx

xxx

xx

3,5

035

xx

xx

3 5

Open Circle

FALSE 81000

81000

151000

FT TFALWAYS graph the TRUE sections…


An application for using intervals and inequalities is figuring out what you would need on a test to achieve a final grade.



Let’s say you have the following test scores and want a “B” for a final grade.

Test scores = 74%, 83%, and a 85%.

What range of scores will earn that final grade of a “B” ?






First, let’s get a total of your test scores : 74 + 83 + 85 = 242







Since there will be four tests used to calculate the final grade, 400 points is the total possible points that could be earned.








So far, this is the equation we have adding that final test and the possible points :









If we need ATLEAST an 80% for a “B”, then the equation needs to be greater than or equal to 80% or 0.80









Now let’s solve for “x”










Multiplied both sides by 400











Subtracted both sides by 242











Subtracted both sides by 242

So the range of scores for a “B” = [ 78 ,100 ]

special equations : and / or and quadratic inequalities and / or are logic operators. and – where...

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