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

Q: Why do we multiply the indices instead of adding them?

Think of it as exponential rules! $ \sqrt[6]{\sqrt{2}} $ means $ (2^{1/2})^{1/6} = 2^{(1/2) \times (1/6)} = 2^{1/12} $, which is $ \sqrt[12]{2} $.

Q: How do I remember which index goes where?

The outer root (6) comes first, then the inner root (2). So you get 6 × 2 = 12. Think of it like peeling an onion - work from outside to inside!

Q: What if there are more than two nested radicals?

Just keep multiplying! For $ \sqrt[3]{\sqrt[4]{\sqrt[5]{x}}} $, you get $ \sqrt[3 \times 4 \times 5]{x} = \sqrt[60]{x} $. Always multiply all the indices together.

Q: Can I convert to exponential form to check my answer?

Absolutely! $ \sqrt[6]{\sqrt{2}} = (2^{1/2})^{1/6} = 2^{1/12} = \sqrt[12]{2} $. Converting to exponential form is a great way to verify your work.

Q: What if the inner radical has a different number under it?

The same rule applies! For $ \sqrt[4]{\sqrt[3]{5}} $, you get $ \sqrt[4 \times 3]{5} = \sqrt[12]{5} $. The base number doesn't change - only the indices multiply.

Nested Radicals with Index Multiplication

Solve the following exercise:

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:06 Let's start by solving this math problem.

00:09 A regular root is the same as a square root.

00:16 Imagine a number, A, under a root of order B, inside another root of order C.

00:22 The result is A raised to the power of B divided by C.

00:28 Let's use this formula to solve our problem.

00:32 Now, calculate the order of multiplication.

00:35 And that's how we find the solution!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve the following exercise:

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

Step-by-step solution

In order to solve this problem, we must simplify the following expression $\sqrt[6]{\sqrt{2}}$ 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 $\sqrt[6]{\sqrt{2}}$ .
Step 2: Recognize that the inner root, $\sqrt{2}$ , can be expressed as $\sqrt[2]{2}$ .
Step 3: Visualize $\sqrt[6]{\sqrt{2}}$ as $\sqrt[6]{\sqrt[2]{2}}$ .
Step 4: Apply the rule $\sqrt[n]{\sqrt[m]{a}} = \sqrt[n \times m]{a}$ .
Step 5: Multiply the indices: $6 \times 2 = 12$ .
Step 6: Replace the compound root with the single root: $\sqrt[12]{2}$ .

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

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

Final Answer

$\sqrt[12]{2}$

Key Points to Remember

Essential concepts to master this topic

Rule: Root of a root equals single root with multiplied indices
Technique: $\sqrt[6]{\sqrt[2]{2}} = \sqrt[6 \times 2]{2} = \sqrt[12]{2}$
Check: Both forms represent the same value: $2^{1/12}$ ✓

Common Mistakes

Avoid these frequent errors

Adding indices instead of multiplying them
Don't add the indices like 6 + 2 = 8 to get $\sqrt[8]{2}$ ! This gives a completely different value that's much larger than the correct answer. Always multiply the indices: 6 × 2 = 12 for $\sqrt[12]{2}$ .

Practice Quiz

Test your knowledge with interactive questions

Solve the following exercise:

$\sqrt[10]{\sqrt[10]{1}}=$

FAQ

Everything you need to know about this question

Why do we multiply the indices instead of adding them?

Think of it as exponential rules! $\sqrt[6]{\sqrt{2}}$ means $(2^{1/2})^{1/6} = 2^{(1/2) \times (1/6)} = 2^{1/12}$ , which is $\sqrt[12]{2}$ .

How do I remember which index goes where?

The outer root (6) comes first, then the inner root (2). So you get 6 × 2 = 12. Think of it like peeling an onion - work from outside to inside!

What if there are more than two nested radicals?

Just keep multiplying! For $\sqrt[3]{\sqrt[4]{\sqrt[5]{x}}}$ , you get $\sqrt[3 \times 4 \times 5]{x} = \sqrt[60]{x}$ . Always multiply all the indices together.

Can I convert to exponential form to check my answer?

Absolutely! $\sqrt[6]{\sqrt{2}} = (2^{1/2})^{1/6} = 2^{1/12} = \sqrt[12]{2}$ . Converting to exponential form is a great way to verify your work.

What if the inner radical has a different number under it?

The same rule applies! For $\sqrt[4]{\sqrt[3]{5}}$ , you get $\sqrt[4 \times 3]{5} = \sqrt[12]{5}$ . The base number doesn't change - only the indices multiply.

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