Solve the Nested Radical: Simplifying the Sixth Root of Square Root of 2

Question

Solve the following exercise:

26= \sqrt[6]{\sqrt{2}}=

Video Solution

Solution Steps

00:00 Solve the following problem
00:03 A 'regular' root is of the order 2
00:10 When we have a number (A) in a root of order (B) in a root of order (C)
00:15 The result equals the number (A) to the power of their quotient (B divided by C)
00:20 We will apply this formula to our exercise
00:25 Let's calculate the order of multiplication
00:29 This is the solution

Step-by-Step Solution

In order to solve this problem, we must simplify the following expression 26 \sqrt[6]{\sqrt{2}} using the rule for roots of roots. This rule states that a root of a root can be written as a single root by multiplying the indices of the radicals.

  • Step 1: Identify the given expression 26 \sqrt[6]{\sqrt{2}} .

  • Step 2: Recognize that the inner root, 2\sqrt{2}, can be expressed as 22\sqrt[2]{2}.

  • Step 3: Visualize 26 \sqrt[6]{\sqrt{2}} as 226 \sqrt[6]{\sqrt[2]{2}} .

  • Step 4: Apply the rule amn=an×m\sqrt[n]{\sqrt[m]{a}} = \sqrt[n \times m]{a}.

  • Step 5: Multiply the indices: 6×2=126 \times 2 = 12.

  • Step 6: Replace the compound root with the single root: 212\sqrt[12]{2}.

Thus, the expression 26 \sqrt[6]{\sqrt{2}} simplifies to 212 \sqrt[12]{2} .

Therefore, the solution to the problem is 212 \sqrt[12]{2} .

Answer

212 \sqrt[12]{2}