1 Roots of Real Numbers: problems block 2 Standard 12 Radical Expressions: problems RADICAL MANIPULATION AND RATIONAL EXPONENTS Rational Exponents: problems

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

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

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:



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.



4

STANDARDS

X4

RADICAL

RADICANDINDEX

This indicates the principal fourth root of X



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…



6

STANDARDS

1

1

1x1 =12

1

= 1

What is the area of the square?

1 = 1

What is the length of the side?



7

2x2 = 22

2

243

21

= 4

4 = 2



STANDARDS



8

3x3 = 32

3

3

987

654

321

=9

9 = 3



STANDARDS



9

4x4 = 42

4

4

16151413

1211109

8765

4321

= 16

16 = 4



STANDARDS



10

5x5 = 52

5

5

2524232221

2019181716

1514131211

109876

54321= 25

25 = 5



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



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



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



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



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…



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?



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!



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



20


a = bn a a a a = b.... n b = a

n

EVEN

b < 0, negative

in b




21


a = bn a a a a = b.... n b = a

n

ODD

b > 0, positive

n b+

None negative

one




22


a = bn a a a a = b.... n b = a

n

ODD

b < 0, negative

n b–

None positive

one




23


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




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.



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:



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



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



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



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:



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



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?



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:



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:



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.



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.



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




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.



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.


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:


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.


42

Standard 12

6416

-= 26

16

-

= 26

16

-

= 2-1

= 12

218747

-= 37

47

-

= 37

47

-

= 3-4

= 134

= 181

Simplify these expressions:


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:


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:


Documents

1 Roots of Real Numbers: problems block 2 Standard 12 Radical Expressions: problems RADICAL MANIPULATION AND RATIONAL EXPONENTS Rational Exponents: problems