Math 6 Honors ( Periods 1, 2, & 3)

The Tangram Story
(One of the activities we did while students were at Astro Camp)

Once upon a time, long, long ago, in a faraway magical land lived a little boy named Tan. The emperor of the land had entrusted Tan with a very special task. The emperor had given Tan a magical square tile and asked him to deliver it to one of the emperor’s subjects who lived in the countryside. Tan was instructed to go directly to the subject’s house and not stop along the way. But, along the way, Tan encountered a group of his friends playing along the river. Tan thought he would stop for just a few minutes to rest and play … when alas; the tile flew out of his pocket and broke into seven pieces. Tan and his friends were so upset that they tried for days to put the tile back together again. They were able to form many beautiful designs of birds and animals and flowers but never the square tile. His designs have been handed down form generation to generation for over 3000 years and are known as Tangram puzzles.
If you would like to enjoy creating some of these puzzles or are interested in how to create your own tile, come in before or after school.

Math 6 Honors ( Periods 1, 2, & 3)

Proportions 7-7
We used fictional numbers for the middle schools in our district
AEWright 6th grade 160 students and 10 teachers
ACStelle 6th grade 144 students and 9 teachers
We looked at the teacher to student ratios of both schools and found
10/160 = 1/16
and
9/144 = 1/16

so 10/160 = 9/144
An equation that states two ratios are equal is called a proportion
10, 160, 9, and 144 are called the terms of the proportion.
Let's say LCMS has 192 students. How many teachers will be needed if the teacher- strudent ratio needs to be the same?

Let t = the number of teachers
10/160 = t/192

Using cross multiplying we discovered that
160t = 192 (10)

160t = 192 (10)
160 160

t = 192/16
or t = 12

12 teachers

we found that when
a/b = c/d and b is not equal to 0 and d is not equal to zero
then
ad = bc


3/8 = 12/n
3n = 8 (12)
3n = 96
3n/3 = 96/3
n = 32

what about 4/3 = n/7
3n = 4(7)
3n = 28

3n/3 = 28/3

n = 9 1/3

3/75 = 2/n
n = 50

3/m = m/27
m² = 3 (27)
m² = 81
so m = 9

49/n = n/4
n² = 4(49)
n² = 196
n = 14

Algebra Honors (Period 6 & 7)

Direct & Inverse Variations 8-9 & 8-10

f(x) = mx + b
It is a linear function. the f(x) is dependent on the x value.
Several examples were given in class:
Truck rental Company charges $35 a day plus 21 cents per mile. Normally your questions in the past were "What is the cost for a rental of a truck for ...:
1 day and 340 miles?" or
"2 days and 450 miles?"
You would just plug in and figure out the exact cost...

Then we discussed the rental of a chain saw ( from the textbook)... remember it is for cutting down trees... like those which were knocked down by those tremendous winds we had a few months back...

We are given that the rental is $5.90 a hour and you must pay $6.50 for 1 can of gas. Again, in the past your questions would be something like...
"How much would it cost for 7.5 hours?"
But... you could write a linear function to represent the cost for all different rental hours.
Let h represent the hours
s(h) = 5.9h + 6.5
Now, no matter how many hours you rent the chain saw, you can figure out the cost.

Phone bills in the past charged 15 cents per message + a base charge. Let's say you were given the July bill of $18 which included 62 messages. What was the base charge?
First find out the cost of the messages (.15)(62) = $9.30 and subtract that from $18.
Or just realize it would be 18 - (.15)(62) = $ 8.70

Then you could write a linear function p(x) = .15x + 8.7

August had 76 messages... what was the bill becomes easy to solve-- just use the linear equation.
p(76) = .15(76) + 8.7 = 20.10. August bill was $20.10

Direct Variation

Direct variation is a function ( abbreviate fcn) defined by the following equation:
y = kx where k is a non-zero constant
y varies directly as x

k is called the constant of variation
( In graphing k represents the slope)

Example:
m varies directly as n
m = 42 and n= 2
Find:a) the constant of variation
b) value of m when n = 3

let m = kn
Finding: a) the constant of variation
42= k(2)
42 = 2k
solve this one step equation
k = 21

finding b) value of m when n = 3
just substitute in
m = (21)(3)= 63

Could we have found this a different way? YES..
notice we have an ordered pair (n,m)
Think about it first in terms of x and y
That is, each pair would be (x₁, y₁) or (x₂, y₂)
if y =kx we have y₁= kx₁ and y₂= kx₂ which means that
y₁)/x₁ = k
and
y₂)/x₂ = k
so we can set the two equal to each other


y₁)/x₁= y₂)/x₂
k is the constant of proportionality
y is directly proportional to x... and we can solve using proportions
So... back to the question
m varies directly as n
m = 42 and n= 2
Find:a) the constant of variation
b) value of m when n = 3

42/2 = x/3
solving using your knowledge of proportions... from 6th grade..
x = 6

We reviewed a few equations and found the following to be direct variations:
y = 3x
p = 9s
d = 3.3t
even y/x = -5
But the following were determined NOT to be direct variatins:
y = 3x²
xy = 4

Eample:
y varies directly as x
y = 6 and x = 72 Find the constant of variation
y = kx
6 = k(72)
6 = 72k
k = 1/12

Turn to Page 394
#20
distance on a map varies directly to actual distance
m= distance on the map
d= actual distance
m = kd
Given that 1 in on the map ---> 10 miles
1 = k(10)
1 = 10k
k = 0.1
so formula is m = 0.1d
writing as a proportion you would have
1/10 = m₂ / d₂

# 22 Volume directly proportional to temp T in Kelvin
5 Liters 300 degrees
V = kT
5= 300k
k = 1/60
so formula is V = (1/60)T
and as a formula before you simplify
5/300 = V₂ /T₂
or 1/60 = V₂ /T₂

Algebra Honors (Period 6 & 7)

Rise Up... Run out.....

Algebra Honors (Period 6 & 7)

Functions Defined by Equations 8-7
A relation is a set of ordered pairs such as
{ (2,3), (3,5), (-4, 0), (5, 0)}
it is also a function because no repeating of the x value,


Domain of a relation is the set of 1st coordinates
the x values
Range is the 2nd coordinates
the y values

So in the example above
Domain is {2, 3, -4}
Range {3, 5, 0}

A relation that assigns to each value in the domain exactly one value in the range is called a FUNCTION
{ (2,3), (3,5), (-4, 0), (5, 0)} is a FUNCTION, whereas,

{ (2,3), (2,5), (-4, 0), (5, 0)} is NOT a function

values of domains ( x's) each are paired with only one element in the range.
Several ways to check... we looked at T- tables to compare the x values, we looked at mapping and we looked at graphs. Notice the vertical line test.

WE then compared the notation for functions
first we looked at y = 3x + 4 vs f(x) = 3x + 4

solve for x = 5
For y = 3x + 5
y = 3(5) + 4
y= 15 + 4
y = 19

Now, what was x again
Oh yeah... x = 5
so the ordered pair is (5, 19)
With f(x) = 3x + 4 however we have
f(5) = 3(5) + 4
f(5) - 15 + 4
f(5) = 19
and you can see what x was originally
(5, 19) is the ordered pair

a
Two ways to show functions
f(x) ... and we noted tht it could be g(x) or h(t) etc
or
f:x-->

g:x-->4 + 3x - x²
if the domain D = { -1, 0, 1, 2}
g:-1--> g(-1) = 4 + 3(-1) - (-1)²= 0
g:0-->g(0) = 4 +3(0) -0 = 4
g:1--> g(1) 4 + 3(1) - (1)² = 6
g:2-->4 + 3(2) - (2)²

Range = {0. 4. 6} Notice that we list each number once ( even if there is a repeated number)

f:x --> x² - 2x for the set of all REAL numbers
find
f(4) = (4)² -2(4) = 8
f(-3) = (-3) ² -2(-3) = 15
f(2) = (2)² -2(2) = 0

so the Range is {8, 15, 0}
Relations & Functions
RELATIONS: Set of ordered pairs where the x values are the DOMAIN and the y values are the RANGE.

FUNCTIONS: Relations where there is just one y value for each x value IN OTHER WORDS----YOU CAN'T HAVE TWO y VALUES for the SAME x value!!!
If you see x repeated twice, it's still a relation, but it's not a function.
In the real world, there are excellent examples....pizza prices.
A restaurant can't have two different prices for the same size cheese pizza.
If you charge $10 and $12 on the same day for the same pizza, you don't have a function.
But, you certainly can charge $10 for a cheese pizza and $12 for a pepperoni pizza.

VERTICAL LINE TEST: When you graph a function, if you draw a vertical line anywhere on the graph, that line will only intersect the function at one point!!!!
If it intersects at 2 or more, it's a relation, but not a function.
So a horizontal line function, y = 4, is a function, but a vertical line function, x = 4 is not.

Any line, y = mx + b, is a function.

INPUTS: x values
OUTPUTS: y values

f(x) means the value of the function at the given x value
You can think of f(x) as the y value

Finding the value of a function: Plug it in, plug it in!
f(x) = 2x + 7
Find f(3)
f(3) = 2(3) + 7 = 13
The function notation gives you more information than using y
If I tell you y = 13 you have no idea what the x value was at that point
But if I tell you f(3) = 13, you know the entire coordinate (3, 13)

Domain of a function = all possible x values (inputs) that keep the solution real
Range of a function = all possible y values (outputs) that result from the domain

EXAMPLE:
f(x) = x + 10 has the domain of all real numbers and the same range because every value will keep the answer f(x) a real number

EXAMPLE:
f(x) = x² has the domain again of all real numbers, BUT the range is greater than or = to zero
because when a number is squared it will never be negative! So f(x) will always be 0 or positive

EXAMPLE:
f(x) = absolute value of x has the domain of all real numbers, but again the range will be greater than or equal to zero because absolute value will never be negative

EXAMPLE:
f(x) = 1/x has a domain of all real numbers EXCEPT FOR ZERO because it would be undefined if zero was in the denominator. The range is all real numbers except zero as well.
This function will approach both axes but never intersect with them.
The axes are called asymptotes which means that they will get very close but never reach them

EXAMPLE:
f(x) = (x - 10)/x + 3

Domain is all real numbers EXCEPT -3 because -3 will turn the denominator into zero (undefined)
What is the range?

Math 6 Honors ( Periods 1, 2, & 3)

Ratios 7-6

In our textbook, the example given involves the number of students --at what I called a mythical middle school --as well as the number of teachers. There are 35 teachers and 525 students. We can compare the number of teachers to the number of students by writing a quotient


number of teachers
number of students

35
525

1/15

The quotient of one number divided by a second number is called the ratio of the first number to the second number.

We can write a ratio in the following ways:

1/15 OR 1:15 OR 1 to 15

All of these expressions are read one to fifteen.

If the colon notation is used the first number is divided by the second. A ratio is said to be lowest terms if the two numbers are “relatively prime.”
You do not change an improper fraction to a mixed number if the improper fraction represents a ratio


There are 9 players on a baseball team. Four of these are infielders and 3 are outfielders. Find each ratio in lowest terms.

a. infielders to outfielders

b. outfields to total players

# of infielders
# of outfielders
= 4/3 or 4:3 or 4 to 3



# of outfielders
# total of players
= 3/9 = 1/3 or 1:3 or 1 to 3

Some ratios compare measurements. In these cases we must be sure the measurements are expressed in the same units

It takes Kiana (or Helen or Emme) 4 minutes to mix some paint. It takes her 3 hours to complete painting her room. What is the ratio of the time it takes Kiana (or Helen or Emme) to mix the paint to the time it takes her to paint her room?
Use minutes as a common unit for measuring time. You must convert the hours to minutes first

3h = 3 • 60min = 180 min

The ratio is :
min. to mix
min. to paint

= 4/180 = 1/45 or 1:45

Some ratios are in the form

40 miles per hour or 5 pencils for a dollar
“ I want my… I want my…. I want my … MPG!!”

These ratios involve quantities of different kinds and are called rates. Rates may be expressed as decimals or mixed numbers. Rates should be simplified to a per unit form. When a rate is expressed in a per unit form, such a rate is often called a unit rate.

I know you will be driving in a blick of an eye... so Justin, in his Lamborghini(and Shane in his Corvette and Nick in his yellow Lamborghini) went 258 miles on 12 gallons of gas. Express the rate of fuel consumption in miles per gallon.

The rate of fuel consumption is

258 miles
12 gallons

= 21 1/2 miles per gallon

Some of the most common units in which rates are given are the following:
mi/gal or mpg miles per gallon
mi/h or mph miles per hour
km/L kilometers per liter
km/h kilometers per hour

Page 229
1 What is the cost of grapes in dollars per kilogram if 4.5 kg of grapes costs $7.56?
$7.56/4.5 kg divide carefully and you discover it is $1.68/kg
2. THe index of refraction of a transparent substance is the ratio of the speed of light in space to the speed of light in the substance.
Using the table from the textbook (look at page 229) Find the index of refraction of
a) glass
300,000/200,000 straight from the chart, which can simplify to 3/2
b) water
300,000/225,000 again from the chart, which can simplify to 4/3

3. The mechanical advantage of a simple machine is the ratio of the weight lifted by the machine to the forse necessary to lift it.
What is the mechanical advantage of a jack that lifts a 3200 pound car with a force of 120 pounds?
3200/120 = 80/3

4. The C string of a cello vibrates 654 times in 5 seconds. How many vibrations per second is this?
654 vibrations/5seconds... divide carefully and you find... 130 4/5 vibrations per second

5. A four-cubic-foot volume of water at sea level weights 250 pounds. What is the density of water in pound per cubic foot?
250 pounds/4 cubic ft ... divide carefully and you find 62 1/2 lb/ft³

6. A share of stock that costs $88 earned $16 last year. What was the price to earnings ratio?
88/16 = 11/2

7. we did in our spiral notebooks this year... please check

Algebra Honors (Period 6 & 7)

Check this out... and have fun learning the words...y = mx +b

Math 6 Honors ( Periods 1, 2, & 3)

Multiplication of Fractions 7-3

If a rectangle is divided into 4 equal parts, each part is ¼ of the whole. If each of these parts is then divided into 3 parts, that is into thirds, then there are 12 equal parts and each is 1/(3 ∙4) or 1/12 of the whole.

That is 1/3 of 1/4 is 1/(3 ∙4) or 1/12 and 1/3 ∙ 1/4 = 1/12 is

so another example 2/3 of 4/5 is 2∙4 /(3∙8) or 2/3 ∙4/5 = 8/15


Notice, that the numerator of the product, 8, is the product of the numerators 2 and 4. The denominator of the product, 15, is the product of the denominators 3 and 5

Rule
If a, b, c, and d are whole numbers with b ≠ 0 and d ≠ 0 , then

a/b(c/d) = a∙c/(b∙d)


When multiplying two fractions, you can simplify the multiplication by dividing either of the numerators and either of the denominators by common factors

6/35 ( 7/3) we can simplify first because both 6 and 3 are divisible by 3
2/35 (7/1) and then both 35 and 7 are divisible by 7 so 2/5 (1(1) = 2/5

Try the following

25/6 ( 42/5) What can we do there?

7/8(20/21) How about with these two sets of fractions?

19/20 ( 25/38) … and these fractions?

What happens when you have
15/2(7/8- 5/24)
What must we do first?

PEMDAS... in my classroom...
15/2( 21/24 - 5/24)
= 15/2(16/24)
= 15/2(2/3)
then simplify to
15/1(1/3)
= 5

What about
8/9∗ 15/32∗ 9/10 = 3/8

or 16/11 × 33/20 × 5/3 = 4

Division of Fractions 7-4

Certain numbers when multiplied together have the product 1
5 X 1/5 = 1
3/4 X 4/3 = 1

Two numbers whose product is 1 are called reciprocals of each other.
Thus 3/4 is the reciprocal of 4/3.
Zero does not have a reciprocal


Look at the following:
We know 18 = 3 X 6 and we know 18 ÷ 6 = 3 as well as 18 X 1/6 = 3
Dividing a number by a fraction is the same as multiplying the number by the RECIPROCAL of the fraction
a/b ÷ c/d = a/b ÷ d/c
Remember- you are using the reciprocal of the divisor... that is , as students want to say "You FLIP the 2nd number!!"

42/ 55 ÷ 36/11
you must rewrite the problem using the reciprocal of the 2nd number
42/55 X 11/36
Now using your skills of observing GCF simplify before you multiply ( MUCH EASIER and FASTER)
42/ 5 X 1/36 which becomes 7/5 X 1/ 6 = 7/30

Algebra Honors (Period 6 & 7)

Points, Lines, and Their Graphs 8-2

We reviewed graphing or plotting an ordered pair as a point on a coordinate plane.
Horizontal axis is the x-axis
vertical axis is the y-axis

origin is at (0,0)

an ordered pair (3,2) lists the coordinates of a point. In this instance we called the Point A
3 is the x-coordinate also know as the abscissa of A
2 is the y-coordinate also known as the ordinate of A

the x- and y-axes are also called coordinate axes and the number plane is often called the coordinate plane. The coordinate axes separate a coordinate plane into four quadrants identified by Roman Numerals. See page 354 for details.
Points on the coordinate axes are NOT considered to be in any quadrant.

The graph of an equation in two variables consists of all the poins that are the graphs of the solutions of the equations.
x + 2y = 6 has the following ordered pairs:
(0,3)
(2,2)
(4,1)
(6,0)
There are infinite number of solutions-- such as
(-2,4)
(1, 2.5)
The graph of all the solutions lie on the straight line that is drawn when the points are connected.

x + 2y = 6 is a linear equation because its graph is a line.
All linear equations in the variables x and y can be written in the form
ax + by = c
or
Ax + By = C
where a, b, and c are real numbers with a and b noth both zero. If a, b, and c are integers, then the equation is said to be in standard form.

2x -5y = 7 and 4x + 9y = 0 and y = 3 are examples of linear equations in standard form

(1/2)x + 4y = 12 is not
y = 3x -1 is not
neither is x²y + 3y = 4
nor xy = 6

Although you only need two points to determine a line, I suggest you plot 3-- whenever possible to guard against mistakes.

The easiest solutions to find are those where the line crosses
the x-axis ( y = 0) and
the y-axis ( x = 0)

Algebra Honors (Period 6 & 7)

Fractional Exponents

In chapter 4 we reviewed the law of exponents:
a^m ⋅aⁿ = a^m+n
Thus you know
2⁴⋅2⁵= 2⁹
What do you notice? What would be the value of n in the equation
2ⁿ⋅2ⁿ = 2
Using what we know from above,
2ⁿ⋅2ⁿ = 2ⁿ⁺ⁿ = 2²ⁿ

The bases are equal ( and NOT -1, 0 or 1). Therefore the exponents must be equal.
That says
2n = 1
n = 1/2

and you have
2^1/2⋅2^1/2=2
Because √2⋅√2 = 2 and (-√2)(-√2) = 2 we note that 2^1/2 as either the positive or negative square root of 2
Selecting the positive or principal square root we define,
2^1/2 = √2

Radicals are not restricted to square roots. The symbol ∛ represents the third ( or cube) root, ∜ represents the fourth root and so on...
As you have learned the root index is omitted when n = 2

Just as the inverse of squaring a number is finding the square root, the inverse of cubing a number is finding the cube root. Since 2³ = 8
∛8 ( read the cube root of 8) is 2.
Likewise (-2)³ = -8
∛(-8) = -2

BE CAREFUL---> While ∛-8 is a real number √-8 is not
In general, you CAN find ODD roots of negative numbers but not EVEN Roots!!

Solve
4ⁿ⋅4ⁿ⋅4ⁿ= 4
4³ⁿ = 4
Since the bases are EQUAL ( that's the KEY), the exponents are also!!
so 3n = 4
n = 3/4


You know that ∛7 = 7 ^1/3 So How would you write (∛7) ² in exponential form?
(∛7) ² = (7^1/3)² = 7^(1/3)2 = 7^2/3

Simplify:

16^3/4
First write as
∜16³
Now change 16 into 2⁴ Why?
You end up with ∜(2⁴)³
Looking at just ∜2⁴ you realize you have 2
and so you are left with
2³ = 8

Math 6 Honors ( Periods 1, 2, & 3)

Addition & Subtraction of Mixed Numbers 7-2

To add or subtract mixed numbers we could first change the mixed numbers to improper fractions and then use the method from 7-1 .
1 4/9 + 3 1/9 = 13/9 + 28/9 = 41/9 = 4 5/9 but that was 5th grade….
In the second method, and the one I prefer, you work separately with the fractional and whole number parts of the given mixed numbers.


STACK THEM!!
3 4/9
1 7/9
4 11/9 = 5 2/9


If the fractional parts of the given mixed numbers have different denominators, we find equivalent mixed numbers whose fractional parts have the same denominator, usually the LCD.

5 3/10 + 7 7/15

Stack

5 3/10
+7 7/15

Draw a line separating the fractional part from the whole numbers Find the LCM of the denominators the LCD and add…

9 5/9 - 4 13/15

Algebra Honors (Period 6 & 7)

Simple Radical Equations 11-10

Solving equations involving radicals are solved by isolating the radical on one side of the equals sign and then squaring both sides of the equation.
140 = √2(9.8)d all under the √
140 = √19.6d
(140)² = (√19.6d)²
19600 = 19.6d
1000=d
The solution set is {1000}
Solve
√(5x+1) + 2 = 6
√(5x+1) = 4
[√(5x+1)]² = (4)²
5x + 1 = 16
5x = 15
x = 3
The solution set is {3}

When you square both sides of an equation, the new equation may NOT be equivalent to the original equation Therefore, you must CHECK EVERY POSSIBLE ROOT IN THE ORIGINAL EQUATION to see whether it is indeeed a root.
Solve
√(11x² -63) - 2x = 0

√(11x² -63) = 2x

√(11x² -63)² = (2x)²
11x² -63 = 4x²
7x² = 63
x² = 9
x = ± 3
Now we need to check for BOTH + 3 and - 3

Rewrite the original equation



√(11x² -63) - 2x = 0
√(11(3)² -63) - 2(3) = 0
√99-63 - 6 = 0
√36 - 6 = 0
6-6 = 0
That's true
Now for x = -3

√(11(-3)² -63) - 2(-3) = 0
√(99 -63) + 6 = 0
√36 + 6 = 0
12 ≠ 0
So -3 is NOT a solution

Math 6 Honors ( Periods 1, 2, & 3)

Addition and Subtraction of Fractions 7-1

Most of you already know how to add and subtract fractions, although some of you may need just a little review.

5/9 + 2/9 = 7/9
13/12 - 5/12 = 8/12 = 2/3
and that
7/9 – 2/9 = 5/9


13/12 - 5/12 = 8/12 = 2/3
a/c + b/c = (a +b)/c where c does not equal 0
a/c - b/c = (a -b)/c

The properties of addition and subtraction of whole numbers also apply to fractions.
If the denominators are the same— add or subtract the numerators AND use the numerator!!

In order to add two fractions with different denominators, we first find two fractions, with a common denominator, equivalent to the given fractions. Then add these two fractions.

The most convenient denominator to use as a common denominator is the least common denominator of LCD, of the two fractions. That is, the least common multiple of the two denominators.

LCD ( a/b, c/d) = LCM(b, d) where b and d both cannot be equal to 0

For example LCD ( 3/4, 5/6) = LCM(4,6) =12

3/4 = 9/12 and 5/6 = 10/12

Let’s do:
7/15 + 8/9

First find the LCD

LCM(15, 9) Do your factor trees or inverted division – or just by knowing!!

15 = 3• 5
9 = 3²

So LCM(15,9) = [every factor to its greatest power] 3²•5 = 45

Then find equivalent factions with a LCD of 45, and add

7/15 = 21/45

8/9 = 40/45

21/45 + 40/45 = 61/45 = 1 16/45

5/6- 11/24
Stack them and use the LCD
5/6 = 20/24
-11/24 = -11/24

9/24 = 3/8
7/12 + 4/9 + 3/4
several strategies ca be used. You can find the LCD for all three you can use the C+ and the A+
and change it to
(7/12 + 3/4) + 4/9
then add the first two factions
7/12 + 3/4 becomes 7/12 + 9/12 = 16/12 = 4/3
then add 4/3 + 4/9
change 4/3 to 12/9
12/9 + 4/9 = 16/9 = 1 7/9

What about 17/10 - ( 3/5 + 5/6)
You must do the parenthesis first
so 3/5 + 5/6
3/5 = 18/30
5/6 = 25/30
43/30
Now you have
17/10 - 43/30
stack those
17/10 = 51/30


51/30
^-43/30
8/30 = 4/15

n + 1/2 = 5/6
you need to isolate the variable so add 1/2 to BOTH sides of the equation
n + 1/2 = 5/6
-1/2 = -1/2 Change 1/2 to 3/6 and subtract carefully
n = 2/6

n = 1/3
make sure to box your answer

Algebra Honors (Period 6 & 7)

Multiplication of Binomials Containing Radicals 11-9

Chapter 5 taught us how to multiply binomials-- we can use those methods when multiplying binomials that contain square root radicals.
(6 + √11)(6 - √11)
The pattern is
(a +b)(a -b) = a² - b²
so using that we get
6² - (√11)²
36 - 11 = 25

Simplify (3 + √5)²
The pattern here is
(a + b)² = a² + 2ab + b²
so ( 3 + √5)² =
3² + 2[(3)(√5)] + (√5)² =
9 + 6√5 + 5 =
14 + 6√5

Simplify (2 √3 - 5√7)²
The pattern here is (a - b)² = a² - 2ab + b²

(2 √3 - 5√7)² =
(2 √3)² -2[(2)(5)(√3)(√7)] +(5√7)² =
4(3) -20√21 +25(7) =
12 -20√21+ 175 =
187 -20√21

If both b and d are nonnegative, then the binomials
a√b + c√d AND a√b - c√d are called conjugates of one another. COnjugates differ ONLY in the sign of one term

if a, b, c, and d are all integers then the product (a√b + c√d)(a√b - c√d) will be an integer... see the first example!!

Conjugates can be used to rationalize binomial denominators that contain radicals.. getting rid of the radicals in the denominator

Rationalize
3/(5- 2√7)

3/(5- 2√7) = [ 3/(5- 2√7)] × [((5+ 2√7)/(5+2√7)]
This doesn't show well here hopefully you can remember what was done in class...
= 3(5 +2√7)/25-(2√7)² =
(15+6√7)/25-28 =
(15+6√7)/-3 =
15/-3 +6√7/-3 =
-5 -2√7



√√