1. 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

2. 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: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 STANDARDS X 4 RADICAL RADICAND INDEX This indicates the principal fourth root of X PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 STANDARDS X 4 + This is the principal fourth root of X. X 4 – This is the opposite of the principal fourth root of X. This represents both fourth roots of X. X 4 + – But, what is a root? Square? Cubic? Etc. Let’s continue to find out… PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 6 STANDARDS 1 1 1x1 = 1 2 1 = 1 What is the area of the square? 1 = 1 What is the length of the side? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 2x2 = 2 2 2 2 43 21 = 4 4 = 2 What is the area of the square? What is the length of the side? STANDARDS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 3x3 = 3 2 3 3 987 654 321 =9 9 = 3 What is the area of the square? What is the length of the side? STANDARDS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 9 4x4 = 4 2 4 4 16151413 1211109 8765 4321 = 16 16 = 4 What is the area of the square? What is the length of the side? STANDARDS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 5x5 = 5 2 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11. 11 2x2 =2 2 2 2 43 21 4 = 4 = 2 3x3 =3 2 3 3 987 654 321 9 =9 = 3 4x4 = 4 2 4 4 16151413 1211109 8765 4321 16 = 16 = 4 1x1 = 5 2 5 5 2524232221 2019181716 1514131211 109876 54321 25 = 25 = 5 THE SQUARE OF A NUMBER 1 1 1x1 =1 2 1 1 = 1 THE SQUARE ROOT OF A NUMBER STANDARDS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 1 1 1 2 2 2 3 3 3 4 4 4 1x1x1 = 1 3 = 1 1 CUBED 2x2x2 = 2 3 = 8 2 CUBED 3x3x3 = 3 3 = 27 3 CUBED 4x4x4 = 4 3 =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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13. Standards RATIONAL EXPONENTS: PREVIEW X a b = X a 1 b a b 2 5 7 3 X 2 5 X 7 3 X 1 2 1 2 Rewrite in radical form: Rational exponents comply with all the exponents’ rules. Simplify the following: = X 1 2 2 3 + 3 3 2 2 3 6 4 6 + 7 6 7 6 2 3 3 5 – 5 5 3 3 1 15 = X 15 = X 10 15 – 9 X 1 2 2 3 X 2 3 X 3 5 X 4 3 5 3 = X 4 3 5 3 20 9 = X 20 9 For any nonzero real number X and integers a and b, with b>1 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. Standards Observe the following pattern: (-1) 2 = (-1)(-1) = +1= +1 (-1) 3 = (-1)(-1)(-1)= –1 (-1) 4 = (-1)(-1)(-1)(-1) = +1= +1 (-1) 5 = (-1)(-1)(-1)(-1)(-1) = –1 (-1) 6 = (-1)(-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: = X 222 = –X 351 = X 876 = –X 999 Let’s continue… PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15. Standards NOW STUDY THE FOLLOWING PATTERN: = 16 = 4 = 16 = 4 = 25 = 5 = 25 = 5 = 36 = 6 = 36 = 6 EVALUATE: X 2 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? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16. 16 Try to find the answer in this experiment: STANDARDS a = 2 a = 4 2 = 2 2 1 2 2 2 2 2 2 2 a = 8 3 = 2 3 1 3 3 3 2 3 3 a = 16 4 = 2 4 1 4 4 4 2 4 4 a = 32 5 = 2 5 1 5 5 5 2 5 5 a = 64 6 = 2 6 1 6 6 6 2 6 6 2 2 2 3 8 = 2 4 16 = 2 5 32 = 2 6 64 = 2 2 4 2 POWERS FACTORS ROOTS HOW? This would be the side of a square or of a cube. square cube EVEN INDEX + POSITIVE RADICAND = POSITIVE ROOTODD INDEX + POSITIVE RADICAND = POSITIVE ROOT PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17. LET’S GO BACK TO OUR POWER TO ROOT SEARCH: STANDARDS a = -2 a = 4 2 a = -8 3 = (-2) 3 1 3 3 3 = -2 a = 16 4 a = -32 5 a = 64 6 3 -8 = -2 5 -32 = -2 2 4 = 2 (-2) 2 2 POWERSFACTORS ROOTSHOW? (-2)(-2) (-2)(-2)(-2) (-2)(-2)(-2)(-2) (-2)(-2)(-2)(-2)(-2) (-2)(-2)(-2)(-2)(-2)(-2) (-2) 3 3 4 16 = 2 = (-2) 5 1 5 5 5 = -2 (-2) 5 5 6 64 = 2 (-2) 4 4 6 6 ODD INDEX + NEGATIVE RADICAND = NEGATIVE ROOT EVEN INDEX + POSITIVE RADICAND = POSITIVE ROOT

18. 18 Now the case where the INDEX is EVEN and the RADICAND NEGATIVE: STANDARDS - 4 = (-1)(4)= (-1) 4 = i 4 = 2 i We have one imaginary root. LET’S PUT ALL TOGETHER IN A TABLE! PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19. 19 SUMMARIZING OF FINDINGS: STANDARDS a = b n a aa = b.... n EVEN b > 0, positive n b + n b – one n b = a POWERS FACTORSROOTS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

20. 20 SUMMARIZING OF FINDINGS: STANDARDS a = b n a aa = b.... n b = a n EVEN b < 0, negative i n b POWERS FACTORSROOTS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

21. 21 SUMMARIZING OF FINDINGS: STANDARDS a = b n a aa = b.... n b = a n ODD b > 0, positive n b + None negative one POWERS FACTORSROOTS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

22. 22 SUMMARIZING OF FINDINGS: STANDARDS a = b n a aa = b.... n b = a n ODD b < 0, negative n b – None positive one POWERS FACTORSROOTS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

23. 23 SUMMARIZING OF FINDINGS: STANDARDS a = b n a aa = 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 POWERS FACTORSROOTS PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

24. 24 Standards LET’S RETAKE THIS PATTERN AGAIN = 16 = 4 = 16 = 4 = 25 = 5 = 25 = 5 = 36 = 6 = 36 = 6 EVALUATE: X 2 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: X 2 = |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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

25. Product Property of Radicals Quotient Product of Radicals Standards MULTIPLYING AND DIVIDING RADICALS: 4 16 = 2 4 = 8 4 16 = 4 16 = 64 = 8 OR 81 9 = 9 3 = 3 81 9 9 = = 9 = 3 In general for n > 1 and any real numbers x and y: x y n = x y n n Exception applies with x < 0 or y < 0 and n even. x y n = x y n n In general for y = 0 and n > 1 as long as all roots are defined: OR Simplify: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

26. 26 Standards Now your turn simplify these using all you learnt before: X 2 = X 2 1 2 2 2 = |X| X 3 = X X 2 1 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 X 4 = X 4 1 2 4 2 2 This is always positive because the EVEN power. X 5 = X X 4 1 4 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

27. 27 Standards LET’S CONTINUE TRYING: X 6 = X 6 1 2 6 2 X 7 = X X 6 1 6 This 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. X 8 = X 8 1 2 8 2 4 X 9 = X X 8 1 8 4 = |X | 3 EVEN ODD needs absolute value to be always positive. = X X 3 3 X 10 = X 10 1 2 = X 10 2 = |X | 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

28. 28 Standards SIMPLIFY THESE: X 11 = X X 10 1 = X X 10 X 12 = X 12 1 2 = X 12 2 =X 6 X 13 = X X 12 1 = X X 12 = X X 6 5 X 14 = X 14 1 2 = X 14 2 = |X | 7 X 15 = X X 14 1 = X X 14 X 16 = X 16 1 2 = X 16 2 =X 8 = X X 7 X 17 = X X 16 1 = X X 16 = X X 8 X 18 = X 18 1 2 = X 18 2 = |X | 9 X 19 = X X 18 1 = X X 18 = X X 9 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

29. 29 Standards X 81 = X X 80 1 = X X 80 X 60 = X 60 1 2 = X 60 2 =X 30 X 73 = X X 72 1 = X X 72 = X X 36 = X X 40 SIMPLIFY: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

30. Standards APPLY IT ALL TOGETHER! 4X Y 2 3 = 4 X Y Y 2 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 Z 10 4 = 5 |Y | Z X 2 5 5 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

31. Standards = X 20 1 4 = X 20 4 = |X | 5 X 20 4 = X 32 1 8 = X 32 8 =X 4 X 32 8 = X 42 1 6 = X 42 6 = |X | 7 X 42 6 = X 80 1 16 = X 80 16 = |X | 5 X 80 16 = X 64 1 16 = X 64 16 =X 4 X 64 16 Can you figure out these? 16X Y 20 21 4 = 16 X Y Y 20 4 4 4 4 = 2 |X | Y Y 5 5 4 5 5 4 Why not absolute value? Why absolute value? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

32. 32 6 81y 2 3 (9y ) = 3 = 9|y | The square roots of 81y are 9|y | 3 6 (x + 2) 8 - 2 4 = - 4 = -(x+2) The opposite of the principal square root of (x+2) is -(x+2) 8 4 35 21 128q r 7 7 3 5 = (2q r ) 7 5 3 = 2q r 35 The principal seventh root of 128q r is 2q r 21 5 3 6 729(y-1) 42 7 = 3 (y-1) 6 6 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 12 Simplify: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

33. 33 Standard 12 x + 10x + 25 2 -5 Since the root index is 2, that is even and the radicand is negative: IT HAS NOT REAL ROOT. 2 2 x + 2x(5) + (5) = 2 = (x + 5) = |x + 5| Simplify: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

34. 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

35. 35 Standard 12 64x q 5 4 2 = 8 (x ) (q ) q 2 2 2 2 2 2 2 2 2 22 = 8x q q 4 5 2401q r 4 4 4 = 7 q q r 4 4 4 4 4 4 4 4 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

36. 36 Standard 12 Simplify this expression: x 6 y 5 (x ) 3 2 (y ) y 2 2 = = 2 2 (x ) 3 2 y y = 3 x y 2 = 3 x 2 2 y = 3 x 2 y = y 3 3 x y rationalizing the denominator There are no radicals left in the denominator. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

37. 37 Standard 12 Simplify this expression: 2 16q 2 7 2 7 2 7 = = 2 q 4 2 7 2 7 3 5 7 3 5 7 = 7 1+3 5 2 q 4+3 5+2 7 2 q 7 7 7 4 5 7 = 16q 5 7 = 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

38. 38 Standard 12 =6 3 3 3 3 - 4 3 3 + 7 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 = 162 3 – 12 3 + 63 3 = 213 3 Simplify the following expression: 6 2187 - 4 27 + 7 243 2187 3 729 3 243 81 27 9 3 1 3 3 3 3 3 243 81 27 9 3 1 3 3 3 3 3 9 3 1 3 3 3 3 2 3 2 3 2 3 2 3 2 3 2 You can see that we did everything to reduce the expression to like terms. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

39. 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 3 2 2 2 2 = 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. 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: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

41. Standard 12 Demonstrate the following equality: 3 – 2 5 2 – 3 2 3 – 2 5 2 – 3 2 = 2 + 3 2 = 3 2 + 3 3 2 – 2 5 2 – 2 5 3 2 2 - 3 2 2 2 = 6 + 9 2 – 4 5 – 6 5 2 4 - 3 2 2 2 = 6 + 9 2 – 4 5 – 6 10 4 - 9(2) = 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

42. 42 Standard 12 64 1 6 - = 2 6 1 6 - = 2 6 1 6 - = 1 2 2187 4 7 - = 3 7 4 7 - = 3 7 4 7 - -4 = 1 3 4 = 1 81 Simplify these expressions: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

43. 43 Standard 12 x 3 5 1 5 = x 3 5 1 5 + 4 5 p 6 7 - = 1 p 6 7 p 1 7 p 1 7 p 6 7 p 1 7 1 7 + = We need to eliminate the rational exponent at the denominator. p p 1 7 = The fraction does not have rational exponent at the denominator, its exponent is now 1. Simplify these expressions: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

44. 44 Standard 12 64 5 16 7 2 5 2 7 6 4 = = 2 2 6 5 4 7 2 6 5 4 7 - = 6 5 4 7 - 7 7 5 5 = 42 35 20 35 - 22 35 = 2 22 35 = Simplify this expression: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved