Simplify the Nested Radical: Solving √⁶(√(25x⁶))

Q: Why do I multiply the exponents 1/2 and 1/6?

When you have nested radicals like $ \sqrt[6]{\sqrt{25x^6}} $, you're applying two operations in sequence. The power rule states $ (a^m)^n = a^{mn} $, so you multiply the exponents: $ \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} $.

Q: How did 5^(1/6) become the 12th root of 25?

Since $ 5 = \sqrt{25} $, we have $ 5^{1/6} = (\sqrt{25})^{1/6} = 25^{1/2 \cdot 1/6} = 25^{1/12} = \sqrt[12]{25} $. This uses the fact that taking a root of a root multiplies the denominators.

Q: Can I simplify this problem in a different order?

Yes! You can use the property $ \sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} $ directly. So $ \sqrt[6]{\sqrt{25x^6}} = \sqrt[12]{25x^6} $, then separate the variables.

Q: Why isn't x^3 the final answer for the x part?

Be careful! We have $ x^6 $ raised to the $ \frac{1}{12} $ power: $ (x^6)^{1/12} = x^{6 \cdot 1/12} = x^{1/2} = \sqrt{x} $. Don't confuse this with $ (x^6)^{1/2} = x^3 $.

Q: Is there a shortcut for nested radicals?

Yes! Remember that $ \sqrt[a]{\sqrt[b]{x}} = \sqrt[ab]{x} $. So $ \sqrt[6]{\sqrt{x}} = \sqrt[6 \times 2]{x} = \sqrt[12]{x} $. This saves steps when working with nested radicals.

Nested Radicals with Fractional Exponents

Complete the following exercise:

$\sqrt[6]{\sqrt{25x^6}}$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve the following problem

00:03 A "regular" root is of the order 2

00:09 When we have a number (A) to the power of (B) in a root of order (C)

00:14 The result equals the number (A) to the power of their product (B times C)

00:18 Let's apply this formula to our exercise

00:27 Let's calculate the product of the orders

00:31 When we have a root of a product (A times B)

00:34 We can write it as a product of the roots of each term

00:39 Let's apply this formula to our exercise, and break down the root

00:47 When we have a number (A) to the power of (B) in a root of order (C)

00:51 The result equals number (A) to the power of their quotient (B divided by C)

00:55 Let's apply this formula to our exercise

01:01 Calculate the quotient of the powers

01:06 Let's apply this formula again in the opposite direction, converting the power to a root

01:16 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Complete the following exercise:

$\sqrt[6]{\sqrt{25x^6}}$

Step-by-step solution

To solve this problem, we'll follow these steps:

Step 1: Convert the expression under the roots to fractional exponents.
Step 2: Simplify the fractional exponents.
Step 3: Return the expression to root form if needed.

Now, let's work through each step:

Step 1: We have the expression $\sqrt[6]{\sqrt{25x^6}}$ . Begin by simplifying the inner square root:

$\sqrt{25x^6} = (25x^6)^{1/2} = 25^{1/2} \cdot (x^6)^{1/2}$ .

We know $25^{1/2} = 5$ and $(x^6)^{1/2} = x^{6 \cdot 1/2} = x^3$ .

So, $\sqrt{25x^6} = 5x^3$ .

Step 2: Apply the outer sixth root:

$\sqrt[6]{5x^3} = (5x^3)^{1/6} = 5^{1/6} \cdot (x^3)^{1/6} = 5^{1/6} \cdot x^{1/2}$ .

Convert $5^{1/6}$ as $\sqrt[12]{25}$ :

Since $5^{1/6} = (5^{1/2})^{1/3} = \sqrt{5}^{1/3} = \sqrt[12]{25}$ ,

we have $5^{1/6} = \sqrt[12]{25}$ .

Therefore, the solution becomes:

$\sqrt{x} \cdot \sqrt[12]{25}$ .

Therefore, the solution to the problem is $\sqrt{x}\cdot\sqrt[12]{25}$ .

Final Answer

$\sqrt{x}\cdot\sqrt[12]{25}$

Key Points to Remember

Essential concepts to master this topic

Rule: Convert nested radicals to fractional exponents first
Technique: $\sqrt[6]{\sqrt{25x^6}} = (25x^6)^{1/2 \cdot 1/6} = (25x^6)^{1/12}$
Check: Verify by converting back to radical form: $\sqrt{x} \cdot \sqrt[12]{25}$ ✓

Common Mistakes

Avoid these frequent errors

Simplifying radicals separately without considering the nested structure
Don't solve $\sqrt{25x^6} = 5x^3$ then $\sqrt[6]{5x^3}$ = wrong approach! This misses the connection between exponents and leads to incomplete simplification. Always multiply the fractional exponents: $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ .

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 I multiply the exponents 1/2 and 1/6?

When you have nested radicals like $\sqrt[6]{\sqrt{25x^6}}$ , you're applying two operations in sequence. The power rule states $(a^m)^n = a^{mn}$ , so you multiply the exponents: $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ .

How did 5^(1/6) become the 12th root of 25?

Since $5 = \sqrt{25}$ , we have $5^{1/6} = (\sqrt{25})^{1/6} = 25^{1/2 \cdot 1/6} = 25^{1/12} = \sqrt[12]{25}$ . This uses the fact that taking a root of a root multiplies the denominators.

Can I simplify this problem in a different order?

Yes! You can use the property $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$ directly. So $\sqrt[6]{\sqrt{25x^6}} = \sqrt[12]{25x^6}$ , then separate the variables.

Why isn't x^3 the final answer for the x part?

Be careful! We have $x^6$ raised to the $\frac{1}{12}$ power: $(x^6)^{1/12} = x^{6 \cdot 1/12} = x^{1/2} = \sqrt{x}$ . Don't confuse this with $(x^6)^{1/2} = x^3$ .

Is there a shortcut for nested radicals?

Yes! Remember that $\sqrt[a]{\sqrt[b]{x}} = \sqrt[ab]{x}$ . So $\sqrt[6]{\sqrt{x}} = \sqrt[6 \times 2]{x} = \sqrt[12]{x}$ . This saves steps when working with nested radicals.

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