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Find the square roots of 9.
3 and – 3
POSITIVE VALUE
5
7

8
 6
 0.12
Simplify
We need to be careful…these are all Principle Square Roots. We
must guarantee positive answers for all of them.
 13
 x  2
 4
 4  4 
 x2
4
16
Factor! Can’t take the
square root unless
there is 1 term.
Can you guarantee
(x – 2) is positive?
x  4x  4
Now it is positive!
x  42
“GAZINTA” RULE
2
x
8
x
4
Divide the root number into
the power on the inside.
2 GAZINTA 8, 4 times.
No absolute value
bars needed.
2
x
6
| x 3 |
2
 x4
10 12 4
x y z
2 GAZINTA 6,
3 times.
| x 5 y 6 z 2 |
Need absolute value bars.
Odd power.
 x y z
5
6 2
ABSOLUTE VALUE
BAR RULE
even
x power  x odd
If an even root simplifies to a
variable with an odd power,
then we need absolute values
around the answer.
 3 3
 2 2
3
2
3
4
2
3
3
2
9
3
 4 4
3
3
3
4
3
2 4
4 16
3
 8  3  2  2
3
2 2
4
“3 GAZINTA 15”
5 times.
“GAZINTA” RULE
 y1
“3 GAZINTA 3”
1 time.
 x2
“3 GAZINTA 6”
2 times.
 z3
“3 GAZINTA 9”
3 times.
3
 27x15
 3x 5
 5  2  2
5
 2 2
5
5
 4 34  3
2 16
4
2
9
3
4
2 2
9
3 3
3
Not Possible to have 4
factors that are the
same equal a negative!
2
 x  1
 y1
“GAZINTA” RULE
6
 x6  6
x6
 x1
 x
  x  4
1
 x4
x 0  1; x  0
x m  x n  x mn
xm
mn

x
xn
x 
m n
Be careful!
Be careful!
Be careful!
4x0  1
4 x 0  4  x 0  4 1  4
2 4  23  4 7
2 4  23  2 7
35
2

1
33
35
2

3
33
 x mn
x
n
1
 n
x
1
m a

x
 
m
x
b
n
b
 
a
n
The numerator is the power
m
n
x  x 
n
m
 x
m
n
The denominator is the root
x
2
3
 2xy   2xy 
1
3
4
2 x y 
1 1
3
3
1
4
1
4
1
4
1
4
2 x y
2
5
 3a b
3
4
4
5
 3 5x 
1
 7
6
1
x
3
13
5 5
3
4
5
 3x 4 y 3
51
6 2
53
6 6
x
2
6
x
x
1
3

13 2 1
4 31
2
y
y
x
1
2

y
 54  12 
 
2 1
3 2
  
2 4 1
5 2
x
y

2 1
3 2
y
x
1
3
2
5
 2x 
20
5
y
 y4
3
6
 2x 
1
2
 a b 
2
 a b 
2
 2x

2 4
 x
4
x
 x
2
1
4
 
 x
 a 8b 4
1
1
4
1
2
x
1
8
8 x
1
3
  x
 x
1
3
12
3
1
4
1
12
 12 x
4
Simplify first then multiply.
The roots must be the same in order
to multiply the radicands together.
 3 2  7  3 14

 3x  4 x  12 x
5
2
5
3
x  2x  2
Can’t simplify
because there are
2 terms inside the
radical.
 x2  4
11 y 4 11y

 
x 5
5x
4
GAZINTA RULE
GAZINTA RULE
2
9
x1 y 2 675y1
8
3 3 2 4
2
2
a6b10 11340a1
20
25 27
2
3 2  2
6 2
5 5 3 9
5  3  x1 y 2 3 y1
15 xy 2
5 9
4
3 3
3y
Remember the Absolute value
bar rule. Even root simplifies
a variable expression to an
odd power!
2
567
2
63
9
2  9  a6b10 5  7a1
18a 6b10 35a
7
GAZINTA RULE
5
27
 xz
1 33
3 3 3
33 5
1500z
2
15 100
3
5 10 10
2 5 5 2
 5 x1 z 3 3 12 z 2
We may want to think of
a way to clean this up!
GAZINTA RULE
Absolute
value bar
rule.
r 3t 2
4
432r1t 3
2 4 27
2r 3t 2
4
27rt 3
Time to clean! Since 432 is
an even number we can agree
that 2 is a factor, but will
there be four factors of 2.
Try 432 divided by 24.
We saw that 27 breaks down
to three 3’s. We need 4, so
don’t break it down.
FACTOR
2
2
GAZINTA RULE


3 x2  2x 1
3x 1x 1
2
3x  1
 h1 5  800 g 4 h 3
2 5 25
2h  5 25 g 4 h3
GAZINTA RULE
Absolute
value bar
rule.
r 3t 4
6
256r 3
26
4
2r 3t 4
6
4r 3
2
x  11 2 3
x 1 3
Remember the Absolute
value bar rule. Even root
simplifies a variable
expression to an odd power!
Since 800 is an even number
we can agree that 2 is a factor,
but will there be five factors of
2. Try 800 divided by 25.
Since 256 is an even number
we can agree that 2 is a factor,
but will there be six factors of
2. Try 256 divided by 26.
Multiply inside with inside, outside with outside! Write as one radical and simplify.
15 6
5 3 3 2
3 10
2  3  3 25  20
5 5 5 4
5 2  3 3 4
303 4
Absolute
value bar
rule.
4
24 10  x 8  y 5
x 2  y1  4 24 10 y1
3 8 2 5
2 2 2
2 x 2 y  4 15 y
All variables will represent positive values in this section!
Simplify first then divide.
80

10
 8 2 2
2 2 2
 53
32
4
5 8
3
 5  2  10
1 72 x 3 y

3
2x
1

36 x 2 y
3
1
  6x y  2x y
3
9 5
18
a
b
4
3a 2b
 4 6a 7 b 4
 ab4 6a 3
Radicals are not allowed in the denominator because they are IRRATIONAL numbers that have
decimals that never end. The PHILOSOPHY: “JUST ENOUGH TO GET BY” finally comes true!
SIMPLIFY FIRST!!

7 22
 2 2
2
2
2
2
2
14
14
 2 
2
4
4

3
2
3
5
1
2 2
2
4
2

2
3 23
2 2 3 22 3
 2 2 
3
3 3


3
3
3

5
24
3
22
3
22
3
20
54


4
4
3
Cube root is under
the MATH button
Break down the denominator into prime factorization.

8x

3
4

8x

2 3  x  y
2
3
1
1
2
3

3
18 x 2 y
21  32  x 2  y1
Multiply by just enough 2’s, 3’s,
x’s and y’s to get groups of 3.
4
8x3 18x 2 y 4 3 18 x 2 y


2  3 x  y
3y
4
24  a 2  b3
5
24  a 2  b3
3y

2 2  31  x1  y 2
5
23  31  x 3  y 2
3y
2 3  x  y
3
1
3
2
112a 2b3c

2ab
5
4

4
54xy2
21  33  x1  y 2
4
54 xy2
3 y 4 54 xy2


2x
2  3 x  y
in simplified form!
Variables must be the same.
Powers on the variables must
be the same.
 15 x  3 y  4
Radicands must be the same.
Roots on the radicals must be
the same.
36
5
16
6 5  2 3 3 5  4 3
3 5 6 3
3
33  2 2 3  2
22  3
5 3x  7 x 2  3x 2  6 3 x
33 2  23 2  2 2
11 3x  4x 2
53 2  2 2
7 xy2
3
2 xy2  12 xy
3
No like radicals.
2 y 2  2x2 y 2
3
2x
 x 3  15
 3 5 2  2  3 2  6
 30  6 3
2 3
 4 3 3  4 38  2  3  2 8
 10 10  3 10  3 10  3  3
222
 12  8 6  6
 87 6
4
 10  3
7
Make a Note!
(a + b)(a – b) = a2 – b2



4
Always use ( )’s around
3  x two or more terms!


4

3x


3x
3x
4 3x
3  x 2 



Multiply top and bottom by
the conjugate of the bottom.
Multiply bottoms 1st.
From the last example…
(a + b)(a – b) = a2 – b2
Bottom doesn’t factor and
nothing can cancel. Done!

3 5  3

 5  3
3 5  3    5  3 
 5  3  5  3
3 5  3  5  3 
53
5 – 3 = 2. FOIL the top and
CLT…looking for a GCF of
2 on the Top.
15
3  5  3 15  15  3
2


18  4 15 2 9  2 15
 9  2 15

2
2
Fractional exponents will be very helpful! Convert.
3
2
 x x
2
3
3
 32  13  22
x
x
1
3
92
6 6
x
 x  x x
11
 a b
2 3
  ab 
1
4
3
4
2
4
We need LCD to add
fractional exponents.
6
11
6
6
5
There is a pattern for
a short cut!
x  x
3
 x
6
3
6
12
2
3
9
12
4
12
8
12
 a  a b b
10
17
12
 a  b 12  12 a10b17
  x  2
 x  2
2 4  33
3 4 4 3
89
12 12
  x  2
17
12
 12 x  2
 x  2 12 x  2
5
4
a 2b3  3 a1b2
3. Multiply the 2nd power to the
1st root and add to the exponent. 12
 6 x11  x  6 x5
  x  2
23
3 4
17
10 5
1. Multiply the roots for a
common root.
2. Multiply the 1st power to
the 2nd root as an exponent.
3
4
2
3
 b a b
1
33 21
1
3
  x  2   x  2
1
3
 a b a b
1 4
3 3
2 4
2 3
3 4
4 3
4 3
 a  a b b3 4
12
2
2
a 2314b3324
 12 a10b17
Fractional exponents will be very helpful! Convert.

x
x
2
2

4
5

41 5
5 2 5
85
10 10
There is a pattern for a short cut!
1. Multiply the roots for a common root.
2
x4
x1
 x
10
 x
85
7
7
4
7
 x
10
3

 5  3  44
  4  44
7
7 3
7 4
 a4 7 b
35  12
21  16
28 28
28 28
a
b
23
5
28
28
 a b
a 5b 3
7
a 3b 4
 a
28
5734
 a
28
b
3512
 x  y 
53
6 4
 x  y 
10  9
 x  y 12 12
25  33
2 6 4 3
3744
 24 x  y 
2116
 24 x  y 
b
 a b
28
3
4
 x  y   12 x  y 1
6
x  y 5
4
x  y 3
 a b
4
x  y 
x  y 
5
6
1
12
23 5
28
4215
2. Multiply the Top power to the
bottom root minus Bottom power
times top root as the exponent.
10
3
7
a b
x
x
3
10
 x  10 x3
5
3
4
a b

1
2
5
4
23 5
5436
2018
 24 x  y   12 x  y 
2
24 and 2 can be reduce by 2,
their GCF.
1
32  32  9  9
33
 3   3
 32   32
FALSE
99
TRUE!
3
3
( x ) (5)
3
x   5 5 5  125
 x   2
 x    7
x  16
WHOA!!!
4
4
4
2
2
Wait a minute.
x  7
Principle Square Roots
are always positive!
NO SOLUTION
NEGATIVE
EVEN
Even root radical are always equal to a positive value. If it is a
negative, then no solution.
+ 17 = +17
2 x 3  8
2
2
x 3  4
POSITIVE…KEEP GOING


x  3  4 
2
x  3  16
+3 = +3
x  19
2
Solve for x.
2 x  5
1
3
3 x  7 1
–7 = –7
2x  5  6  3
3
3 x  6
3
–6 = –6
3
3
2 x  5  3
Cube Roots can be Negative
Keep Going!
x  2
NEGATIVE! NO SOLUTION
 x    2
2
63
2

3

2 x  5   3
3
3
2x  5  27
x4
+5 = +5
2 x  22
2
1
2
x  11
Solve for x. x  x  7  5
–5 =
FACTOR
x 5  x 7
x  2x  9  0
NEGATIVE? I don’t know!
But I know that x > 5 to make it positive!
 x  5
2
x 2  11x  18  0
–5


x7

2
x  5x  5  x  7
F.O.I.L.
x 2  5x  5x  25  x  7
Set = 0
Combine Like Terms
x 2  10 x  25  x  7
–x –7
=–x–7
x 2  11x  18  0
x2 0
x2
9 X
x 9  0
x9
Solve for x.

2x  5  1  x  3
 
2 x  5  1 
2
The radical on the left is ISOLATED so square
both sides…REMEMBER to FOIL the right side!

x  3 1 
2
2x  5  1 x  3
The right side promises to be positive if x > 3.
x 3

2x  5  1  x  3  x  3  x  3
2x  5  2 x  3  x  2
–x +2 =
–x +2
x 3  2 x 3
x  3  2 x  3 
x  3x  3  4x  3
The radical term is isolated. Don’t
divide by 2 it will create fractions.
Square both sides again.
2
2
x 2  3x  3x  9  4 x  12
x  6 x  9  4 x  12
2
– 4x + 12 = – 4x + 12
x  10 x  21  0
x  3x  7  0
Combine Like Terms and
ISOLATE the radical.
2
x 3
REMEMBER to FOIL the left side
and distribute the 4.
Combine Like Terms and set equal to zero.
Factor and solve for x.
x7
Solve for x.
2x  5  1  x  3
 2x  5
Graph and check. Set the equation equal to zero.
 2x  5
Y  0  1  x  3  2x  5
To the calculator. Find the Y= button and press it.
Type in the above equation without the = 0.
Graph in the Standard window by hitting ZOOM 6.
The graph is very close to the x-axis. Hit the TRACE button
to see the x and y coordinates at the bottom of the screen.
In the upper left hand corner will be the equation of the
graph and the coordinates are at the bottom. Notice that
there is no y value. Remember we said that x > to 3. We
won’t see a y until we get to 3.
x 3 x7
Type in a 3 and hit ENTER. Y = 0 means 3 is the right answer.
Type in a 7 and hit ENTER. Y = 0 means 3 is the right answer.
Solve for x.

x5  2 x7
 
x  5  2 
2
The radical on the left is ISOLATED so square
both sides…REMEMBER to FOIL the right side!

x  7 2 
x5  2 x7
2
The right side promises to be positive if x > 7.
x7

x5  42 x7 2 x7  x7
Combine Like Terms and
ISOLATE the radical.
x 5  x 3 4 x 7
–x +3 =–x +3
The radical term is isolated. Divide
by 4 because it divides evenly.
Square both sides again.
8  4 x7
4
4
2  x7
2
2


x7
4  x7
+7 =
11  x
+7

2
Since our answer is 11, it will not fit in the WINDOW. Hit
the WINDOW button and change the Xmax to 15 and hit
GRAPH. TRACE and type in 11 to check it.
a leg
90
hypotenuse
c
leg
b
Remember…positive answers!
c
a2  b2  c2
10 2  90 2  c 2
100  8100  c 2
8200  c 2
a=
b=
8200  c
c  10 82  90.55
Exact Value
Decimal
Value
45 : 45 : 90
b
b:b:b 2
Draw in a line to make a right triangle.
45
b
Solve for c.
45
b
c b 2
a=b
45
a2  b2  c2
b2  b2  c2
2b 2  c 2
45
b
b 2 c
b
Label the sides with our triple ratio.
Solve for b.
b 2 6 2
b
b
45
6
b 2
4 2 b
b 2 8
2
b
4 2
2
8
2
b

2
2
Just found out what b is equal to!
b
8 2
4 2
2
30 : 60 : 90
a : a 3 : 2a
Draw in a line to make a right triangle.
Solve for b.
60
30
30 30
2a
60
2a
b
a
a
c = 2a
60
2
ba 3
60
a 2  b 2  2a 
a 2  b 2  4a 2
b 2  3a 2
ba 3
90
a
2a
Label the sides with our triple ratio.
60
a
 2a  2  8  16
60
4 3 a
30
a 3
8 3
Just found out what a is equal to!
Solve for a.
 2a
8 3
30
a 3
a 3  12
12
3
a

3
3
a
12 3
4 3
3
a
a 3
2a
9 3cm.
18cm.
10 3cm.
5 3cm.
6cm.
4cm.
6 3cm.
8cm.
b
b 2
b
7cm.
7 2cm.
10 2cm.
10cm.
6 2cm.
6 2cm.
8cm.
8cm.
 x  x y  y2 
M 1 2 , 1

2
 2
c= d 
Ax1 , y1 
The distance formula can be derived
from the Pythagorean Theorem.
Build a right triangle.
a b  c
2
2
b  x2  x1 
2
Bx2 , y2 
a   y2  y1 
Cx2 , y1 
 y2  y1 2  x2  x1 2  d 2
 y2  y1 2  x2  x1 2  d
d
x2  x1 2   y2  y1 2
There is no difference on which coordinates are subtracted first. What is important
that when you square the difference it will be positive.
The Midpoint Formula is going to be considered
as the “Average between the two points.”
 x  x y  y2 
M 1 2 , 1

2 
 2
Find the distance and the midpoint between (-3, 5) and (9, -11).
x1 , y1  x2 , y2 
2
2
d
x2  x1    y2  y1 
9   32  11  52
d
122  162
d
 x  x y  y2 
M 1 2 , 1

2 
 2
d  144  256 d  400  20
  3  9 5   11 
M
,

2
 2

6 6
M ,

2 2 
M 3,3
If M(3, 7) is the midpoint of segment AB, find the location of B when A is located
at (-6, 10). Set the Midpoint Formula = (3, 7) and substitute in the coordinates of A.
 x1  x2 y1  y2 
M
,
  3,7 
2 
 2
Short-cut
M 3,7
x1  6
y  10
3 1
7
2
2
*2
Multiply both equations by 2.
 x  6 y1  10 
M 1
,
  3,7  x1  6  6 y1  10  14
2 
+6 =+6
– 10 = – 10
 2
x1  12 y1  4
B12,4
Subtract
the given
endpoint
6,14
A 6,10
B12,4
Given. i   1
i1  i
i 2  1 Notice a pattern!
i 3  i 2  i  1 i  i
i to an odd power is always
equal to i, just need to
determine the sign.
i
i 4  i 2  i 2   1   1  1
i 6  i 4  i 2  1  1  1
i 7  i 4  i 3  1  i   i
n2
i 8  i 4  i 4  1 1  1
 i
 i
i to an even power is always
equal to 1, just need to
determine the sign.
i even  1
Time to determine the sign.
Odd number of negatives multiplied together is
negative and an Even number of negatives is a positive.
Find the number of i2 are in in.
i 5  i 4  i  1 i  i
i 75
odd
i 2  1
i 89
 i
EVEN Whole number = +
ODD Whole number = –
i102 1
75  2  37.5 89  2  44.5 102  2  51
i124 1
124  2  62
Simplify radicals with negatives.
3
 1 3
1  3
i 3
i  5 i  20
i i 5 20
i
4
i 4
i2
2i
i  11
i  50
 i 11
5i 25 2
 5i 2
i  16 i  49
i 16  i 49
i 2 100
4i  7i
 10
11i
Definition of Complex Numbers.
a  bi 
Real
Number
imaginary
Number
Complex Numbers.
Real Numbers.
Imaginary Numbers.
i 2  1
When the directions read, “Leave the answers in a + bi form.” The answer
will have to include a zero if there is no real number or imaginary number.
For example.
If the answer is 2, then we write the answer as 2 + 0i.
If the answer is -5i, then we write the answer as 0 – 5i.
Combine Like Terms. Treat i like a variable.
Distribute the minus sign.
12  2  3i  7i
14  4i
2i  5  2i  3i
10i  6i 2
6  5i  9  8i
6  9  5i  8i
15 13i
4  3  4  2i  5i  3  5i  2i
12  8i  15i  10i 2
3  3  3  2i  2i  3  2i  2i
9  6i  6i  4i 2
10i  61
12  7i 101
 6  10i
12  7i  10
9  41
94
22  7i
13
Complex Conjugate Product Rule.
5  3i 5  3i 
 5 2  32
a  bi a  bi   a 2  b2
 25  9  34
Rationalize the Denominator above rule!
2  3i   5  2i 
5  2i  5  2i 
7  4i   i
 62
10  4i  15i  6i
25  4
7i  4i 
16  11i 16 11

 i
29
29 29
5i
The denominator is a
single term, just multiply
by i top and bottom.
i
2
5i 2
4 7i
 
5 5
7i  4

5

4  7i
5