1 roots of real numbers: problems block 2 standard 12 radical expressions: problems radical...
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1
Roots of Real Numbers: problems block 2
Standard 12
Radical Expressions: problems
RADICAL MANIPULATION
AND
RATIONAL EXPONENTS
Rational Exponents: problems
END SHOW
ROOTS: Which one I’ll get?
INTRODUCTION
Roots of Real Numbers: problems block 1
MULTIPLY AND DIVIDE RADICALS
RATIONAL EXPONENTS VS RADICALS
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2
STANDARD 12:
Students know the laws of fractional exponents, understand exponential functions, and use these functions in problems involving exponential growth and decay.
ALGEBRA II STANDARDS THIS LESSON AIMS:
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3
ESTÁNDAR 12:
Los estudiantes conocen las leyes de los exponentes fraccionarios, entienden funciones exponenciales, y usan estas funciones en problemas que involucran crecimiento exponencial y disminución exponencial.
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4
STANDARDS
X4
RADICAL
RADICANDINDEX
This indicates the principal fourth root of X
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5
STANDARDS
X4+ This is the principal fourth root of X.
X4–
This is the opposite of the principal fourth root of X.
This represents both fourth roots of X.X4+–
But, what is a root? Square? Cubic? Etc. Let’s continue to find out…
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6
STANDARDS
1
1
1x1 =12
1
= 1
What is the area of the square?
1 = 1
What is the length of the side?
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7
2x2 = 22
2
243
21
= 4
4 = 2
What is the area of the square?
What is the length of the side?
STANDARDS
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8
3x3 = 32
3
3
987
654
321
=9
9 = 3
What is the area of the square?
What is the length of the side?
STANDARDS
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9
4x4 = 42
4
4
16151413
1211109
8765
4321
= 16
16 = 4
What is the area of the square?
What is the length of the side?
STANDARDS
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10
5x5 = 52
5
5
2524232221
2019181716
1514131211
109876
54321= 25
25 = 5
What is the area of the square?
What is the length of the side?
The SQUARE OF A NUMBER is the total of square units used to form a larger square.
The SQUARE ROOT OF A NUMBER is the opposite of the square. It is when you find the lenght of the side in a square with a given number of square units.
STANDARDS
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11
2x2 = 22
2
243
21
4
= 4
= 2
3x3 = 323
3
987
654
321
9
=9
= 3
4x4 = 424
4
16151413
1211109
8765
4321
16
= 16
= 4
1x1 = 52
5
5
2524232221
2019181716
1514131211
109876
54321
25
= 25
= 5
THE SQUARE OF A NUMBER
1
1
1x1 =12
1
1
= 1
= 1
THE SQUARE ROOT OF A NUMBER
STANDARDS
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12
11
1
2
2
23
3
3
4
4
4
1x1x1 = 13
= 1
1 CUBED
2x2x2 = 23
= 8
2 CUBED
3x3x3 = 33
= 27
3 CUBED
4x4x4 = 43
=64
4 CUBED
What is the volume for these cubes in cubic units?
3 64 = 4
3 27 = 3 3 8 = 2
3 1 = 1
We can observe that the number to the cube is the volume and the third root of this number is the side of the cube formed with those cubic units.
STANDARDS
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StandardsRATIONAL EXPONENTS: PREVIEW
Xa
b = Xa
1
b= X
ab
= X25
= X73
X2
5
X7
3
X1
2 = X12
= X
Rewrite in radical form:
Rational exponents comply with all the exponents’ rules. Simplify the following:
= X1
22
3+
3
32
2
= X3
6
4
6+
= X7
6
= X76
= X2
33
5–
5
53
3
= X 1
15
= X15
= X10
15– 9
15
X X 1
22
3
X2
3
X3
5
X4
3
5
3
= X4
3
5
3
= X20
9
= X209
For any nonzero real number X and integers a and b, with b>1
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Standards
Observe the following pattern:
(-1) 2 = (-1)(-1) = +1
(-1) 3= (-1)(-1)(-1) = –1
(-1) 4 = (-1)(-1)(-1)(-1) = +1
(-1) 5= (-1)(-1)(-1)(-1)(-1) = –1
(-1) 6 = (-1)(-1)(-1)(-1)(-1)(-1) = +1(-1) 7= (-1)(-1)(-1)(-1)(-1)(-1)(-1) = –1
EVEN powers yield a positive (+) value.
ODD powers yield a negative (–) value.
(–1)EVEN
= +
(–1)ODD
= –
(-X)222
(-X)351
(-X)876
(-X)999
Simplify:
With a negative value:
= X222
= –X351
= X 876
= –X 999
Let’s continue…
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StandardsNOW STUDY THE FOLLOWING PATTERN:
= 16 = 4
= 16 = 4
= 25 = 5
= 25 = 5
= 36 = 6
= 36 = 6
EVALUATE: X2
X = – 4
X = 4
(-4)2
( 4 )2
X = – 5
X = 5
(-5)2
( 5 )2
X = – 6
X = 6
(-6)2
( 6 )2
Observe that for each POSITIVE OUTPUT there are two values of X which are opposite.
e.g. Radicand 36 is from {-6, 6} and index 2, which is EVEN.
May this be generalized for all EVEN INDEXES and POSITIVE RADICANDS?
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16
Try to find the answer in this experiment:
STANDARDS
a = 2
a = 42 = 2 2
1
2= 2
2
2 =22 2 222
a = 83 = 2 3
1
3= 2
3
3 =2233
a = 164 = 24
1
4= 2
4
4 =2244
a = 325 = 2 5
1
5= 2
5
5 =2255
a = 646 = 2 6
1
6= 2
6
6 =2266
2 2 2
2 2 2 2
2 2 2 2 2
2 2 2 2 2 2
3 8 = 2
4 16 = 2
5 32 = 2
6 64 = 2
2 4 = 2
2
POWERS FACTORS ROOTS HOW?
This would be the side of a square or of a cube.
square
cube
EVEN INDEX + POSITIVE RADICAND = POSITIVE ROOT
ODD INDEX + POSITIVE RADICAND = POSITIVE ROOTPRESENTATION CREATED BY SIMON PEREZ. All rights reserved
LET’S GO BACK TO OUR POWER TO ROOT SEARCH:
STANDARDS
a = -2
a = 42
a = -83 = (-2) 3
1
3= (-2)
3
3 = -2
a = 164
a = -325
a = 646
3 -8 = -2
5 -32 = -2
2 4 = 2(-2)22
POWERS FACTORS ROOTS HOW?
(-2)(-2)
(-2)(-2)(-2)
(-2)(-2)(-2)(-2)
(-2)(-2)(-2)(-2)(-2)
(-2)(-2)(-2)(-2)(-2)(-2)
(-2)33
4 16 = 2
= (-2)5
1
5= (-2)
5
5= -2(-2)55
6 64 = 2
(-2)44
(-2)66
ODD INDEX + NEGATIVE RADICAND = NEGATIVE ROOT
EVEN INDEX + POSITIVE RADICAND = POSITIVE ROOT
18
Now the case where the INDEX is EVEN and the RADICAND NEGATIVE:
STANDARDS
- 4 = (-1)(4)
= (-1) 4
= i 4= 2i
We have one imaginary root.
LET’S PUT ALL TOGETHER IN A TABLE!
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19
SUMMARIZING OF FINDINGS: STANDARDS
a = bn a a a a = b....
n
EVEN
b > 0, positive
n b+
n b–
one
one
n b = a
POWERS FACTORS ROOTS
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20
SUMMARIZING OF FINDINGS: STANDARDS
a = bn a a a a = b.... n b = a
n
EVEN
b < 0, negative
in b
POWERS FACTORS ROOTS
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21
SUMMARIZING OF FINDINGS: STANDARDS
a = bn a a a a = b.... n b = a
n
ODD
b > 0, positive
n b+
None negative
one
POWERS FACTORS ROOTS
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22
SUMMARIZING OF FINDINGS: STANDARDS
a = bn a a a a = b.... n b = a
n
ODD
b < 0, negative
n b–
None positive
one
POWERS FACTORS ROOTS
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23
SUMMARIZING OF FINDINGS: STANDARDS
a = bn a a a a = b.... n b = a
n
EVEN
ODD
b > 0, positive b < 0, negative
n b+
n b–
n b+
None negative
i n b
n b–
None positive
one
one
one
one
POWERS FACTORS ROOTS
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24
StandardsLET’S RETAKE THIS PATTERN AGAIN
= 16 = 4
= 16 = 4
= 25 = 5
= 25 = 5
= 36 = 6
= 36 = 6
EVALUATE: X2
X = – 4
X = 4
(-4)2
( 4 )2
X = – 5
X = 5
(-5)2
( 5 )2
X = – 6
X = 6
(-6)2
( 6 )2
Because the power is EVEN the result is ALWAYS POSITIVE
In general:
X2 = |X|
The power is 1: ODD
The absolute value makes sure that the result is always POSITIVE if the power of the radicand is even and the resulting output has an ODD POWER.
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Product Property of Radicals
Quotient Product of Radicals
StandardsMULTIPLYING AND DIVIDING RADICALS:
4 16 = 2 4= 8
4 16 = 4 16
= 64
= 8
OR
81
9=
93 = 3
81
9
819
= = 9 = 3
In general for n > 1 and any real numbers x and y:
x yn = x yn n
Exception applies with x < 0 or y < 0 and n even.
xyn =
x
yn
n
In general for y = 0 and n > 1 as long as all roots are defined:
OR
Simplify:
Simplify:
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26
StandardsNow your turn simplify these using all you learnt before:
X2 = X
2
1
2
= X2
2
= |X|
X3 = X X 2 1
= X X 2
= |X| X
This X can’t be negative for this to yield a real number and therefore X is always positive and the absolute value is redundant.
= X X
X4 = X
4
1
2
= X4
2
=X2
This is always positive because the EVEN power.
X5 = X X 4 1
= X X 4
= X X2
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27
StandardsLET’S CONTINUE TRYING:
X6 = X
6
1
2
= X6
2
X7 = X X 6 1
= X X 6This X can’t be negative for this to yield a real number and therefore X is always positive and the absolute value is not necessary.
X8 = X
8
1
2
= X8
2
=X4
X9 = X X 8 1
= X X 8
= X X4
= |X | 3
EVEN
ODD needs absolute value to be always positive.
= X X3
= X X3
X10 = X
10
1
2
= X10
2
= |X | 5
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28
StandardsSIMPLIFY THESE:
X11 = X X 10 1
= X X 10
X12 = X
12
1
2
= X12
2
=X6
X13 = X X 12 1
= X X 12
= X X6
= X X5
X14 = X
14
1
2
= X14
2
= |X | 7
X15 = X X 14 1
= X X 14
X16 = X
16
1
2
= X16
2
=X8
= X X7
X17 = X X 16 1
= X X 16
= X X8
X18 = X
18
1
2
= X18
2
= |X | 9
X19 = X X 18 1
= X X 18
= X X9
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29
Standards
X81 = X X 80 1
= X X 80
X60 = X
60
1
2
= X60
2
=X 30
X73 = X X 72 1
= X X 72
= X X36
= X X40
SIMPLIFY:
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StandardsAPPLY IT ALL TOGETHER!
4X Y2 3 = 4 X Y Y2 2
= 2 |X| Y Y
This comes from an EVEN power in the radicand and its power is an ODD number; but it doesn’t need the absolute value because the Y as radicand in the radical assures that it is always positive to yield a real number.
25XY Z 10 4 = 25 X Y Z10 4
= 5 |Y | Z X25
= 5|Y |Z X5 2
= 2|X|Y Y
It comes from an EVEN power in the radicand and its power is ODD, therefore it needs the absolute value.
It has an EVEN power and the result is always positive it doesn’t need absolute value
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Standards
= X20
1
4
= X20
4
= |X | 5
X204
= X32
1
8
= X32
8
=X 4
X328
= X42
1
6
= X42
6
= |X | 7
X426
= X80
1
16
= X80
16
= |X | 5
X8016
= X64
1
16
= X64
16
=X 4
X6416
Can you figure out these?
16X Y20 214 = 16 X Y Y20 2044 4 4
= 2 |X | Y Y5 5 4
= 2|X |Y Y5 5 4
Why not absolute value?
Why absolute value?
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32
681y 23(9y )=3= 9|y |
The square roots of 81y are 9|y |3
6
(x + 2)8- 24(x + 2)= -4= -(x+2)
The opposite of the principal square root of (x+2) is -(x+2)8 4
3521128q r773 5= (2q r )7
53= 2q r
35
The principal seventh root of 128q r is 2q r21 53
6 729(y-1)42 7= 3 (y-1)66
6
7= 3|(y-1) |
Note the absolute value is because the resulting power is odd and the root index was even. This prevents the principal root from getting a negative value.
Standard 12Simplify:
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33
Standard 12
x + 10x + 252
-5Since the root index is 2, that is even and the radicand is negative: IT HAS NOT REAL ROOT.
2 2x + 2x(5) + (5)=
2= (x + 5)
= |x + 5|
Simplify:
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34
Standard 12
CONDITIONS TO FULLY SIMPLIFY A RADICAL EXPRESSION:
• The INDEX, is the smallest possible one.
• The RADICAND does not have other factors than 1, which are the nth powers of an integer or polynomial.
• There aren’t any fractions in the denominator.
• There are no radicals left in the denominator.
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35
Standard 12
64x q 54
2= 8 (x ) (q ) q22 22
22= 8 (x ) (q ) q2 2 2
2 2= 8x q q
452401q r 4 44= 7 q q r 44
4= 7 q q r4 44 44 4
= 7 |q| q |r|4
4= 7q|r| q
q and r had absolute value because their indexes were even and their final power was odd. But q by definition has to be positive within the 4th root, therefore, we may remove the absolute value from q, outside the radical, because is redundant.
factor into squares
product property of radicals
factor into powers of 4
product property of radicals
Simplify each expression:
• The RADICAND does not have other factors than 1, which are the nth powers of an integer or polynomial.
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36
Standard 12Simplify this expression:
x 6
y 5
(x )3 2
(y ) y2 2=
=(y ) y2 2
(x )3 2
y
y=
3x
y y2
=3x
y y22
y
=3x
y y2
y=
y3
3x y
rationalizing the denominator
• There are no radicals left in the denominator.
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37
Standard 12Simplify this expression:
216q27
27
16q27
=
=2 q4 27
27
2 q3 57
2 q3 57
=7 2 q1+3 5
2 q4+3 5+27
2 q7 77
2 q4 57
=
2q
16q57
=
To rationalize the denominator we multiply by a factor that give us for each integer or polynomial in the radicand at the denominator the same number as power, as the number we have for the index in the radical.
• There are no radicals left in the denominator.
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38
Standard 12
=6 3 3 3 3 - 4 3 3 + 7 3 3 32 2 2 2 2 2
=6 3 3 3 3 - 4 3 3 + 7 3 3 32 2 2 2 22
= 6 3 3 3 3 - 4 3 3 + 7 3 3 3
= 162 3 – 12 3 + 63 3
= 213 3
Simplify the following expression:
6 2187 - 4 27 + 7 243
6 2187 - 4 27 + 7 243
2187 3729 32438127
931
33333
2438127
931
33333
27931
333
32
32
32
32
32
32You can see that we did everything to reduce the expression to like terms.
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39
Standard 12
Simplify the following expression:
(4 3 - 2 3 )(2 + 4 9 )
(4 3 - 2 3 )(2 – 4 9 ) = 4 3 2 + 4 3 4 9 - 2 3 2 - 2 3 4 9
= 8 3 + 16 3 9 - 4 3 - 8 3 9
F O I L
= 8 3 +16 3 3 - 4 3 - 8 3 32 2
= 8 3 +16 3 3 - 4 3 - 8 3 32 2
= 8 3 +16 3 3 - 4 3 - 8 3 3
= 8 3 +48 3 - 4 3 - 24 3
= 28 3 You can see that we did everything to reduce the expression to like terms.PRESENTATION CREATED BY SIMON PEREZ. All rights
reserved
40
Standard 12
1 – 3 2 1 + 3 2
4 – 3 5 4 + 3 5
2 – 3 2 + 3
5 2 – 1 5 2 + 1
6 + 5 7 6 – 5 7
Complete the conjugates for the following:
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Standard 12Demonstrate the following equality:
3 – 2 5
2 – 3 2
3 – 2 5
2 – 3 2=
2 + 3 2
2 + 3 2
=3 2 + 3 3 2 – 2 5 2 – 2 5 3 2
2 - 3 222
=6 + 9 2 – 4 5 – 6 5 2
4 - 3 222
=6 + 9 2 – 4 5 – 6 10
4 - 9(2)
-1-1=
6 + 9 2 – 4 5 – 6 10 -14
=6 10 + 4 5 - 9 2 - 6
14
3 – 2 5
2 – 3 2=
6 10 + 4 5 - 9 2 - 6 14
To rationalize the denominator we multiply by the conjugate.
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42
Standard 12
6416
-= 26
16
-
= 26
16
-
= 2-1
= 12
218747
-= 37
47
-
= 37
47
-
= 3-4
= 134
= 181
Simplify these expressions:
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43
Standard 12
x x35
15 = x
35
15
+
= x45
p67
-=
1p
67 p
17
p17
p67
p17
17+
=
We need to eliminate the rational exponent at the denominator.
pp
17
=The fraction does not have rational exponent at the denominator, its exponent is now 1.
Simplify these expressions:
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44
Standard 12
645
167
25
27
6
4=
= 22
65
47
265
47-=
65
47
-77
55 =
4235
2035
- 2235
=
22235=
Simplify this expression:
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