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Algebra 2
   32 = 9

   102 = 100

   2172 = 47089
   23 = 8

   53 = 125

   1713 = 5000211
   For an integer n greater than 1,
    if b n = a, then




   The nth root of a is equal to b
Let n be an integer (n > 1), a is a real number and
n is an even integer

   a < 0, no real nth roots

   a = 0, one real nth root

   a > 0, two real nth roots
Let n be an integer (n > 1), a is a real number and
n is an odd integer

   a < 0, one real nth root

   a = 0, one real nth root

   a > 0, one real nth root
   n = 5, a = -32

   Answer, n is odd so there is 1
    solution:

   n = 6, a = 1

   Answer – n is even, a is positive so 1
    solution:
   Rational exponents don’t have to be in the
    form 1/n
   125    2/3




   Answer : 125   2/3   = (125   1/3)2

    = 52 = 25

   8   -4/3


   Answer: 1/16
   Solve: 4x5 = 128


   Solve: (x – 3)4 = 21


   Solve: x6 – 34 = 181
Solve: 4x5 = 128
x = 2


 Solve: (x – 3)4 = 21
 x=


 Solve: x6 – 34 = 181
 x=

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Chapter 6.1

  • 2. 32 = 9  102 = 100  2172 = 47089
  • 3. 23 = 8  53 = 125  1713 = 5000211
  • 4. For an integer n greater than 1, if b n = a, then   The nth root of a is equal to b
  • 5.
  • 6. Let n be an integer (n > 1), a is a real number and n is an even integer  a < 0, no real nth roots  a = 0, one real nth root  a > 0, two real nth roots
  • 7. Let n be an integer (n > 1), a is a real number and n is an odd integer  a < 0, one real nth root  a = 0, one real nth root  a > 0, one real nth root
  • 8. n = 5, a = -32  Answer, n is odd so there is 1 solution:  n = 6, a = 1  Answer – n is even, a is positive so 1 solution:
  • 9. Rational exponents don’t have to be in the form 1/n
  • 10. 125 2/3  Answer : 125 2/3 = (125 1/3)2 = 52 = 25  8 -4/3  Answer: 1/16
  • 11. Solve: 4x5 = 128  Solve: (x – 3)4 = 21  Solve: x6 – 34 = 181
  • 12. Solve: 4x5 = 128 x = 2  Solve: (x – 3)4 = 21  x=  Solve: x6 – 34 = 181  x=