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



√√