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

Video Solution

Solution Steps

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 Solution

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

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