Algebra Honors ( Periods 5 & 6)

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


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

Algebra Honors (Periods 5 & 6)

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



√√

Math 6A (Periods 2 & 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

Math 6A (Periods 2 & 4)

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