Solve Nested Square Roots: √√16 × √√8 Multiplication Problem

Question

Complete the following exercise:

168= \sqrt{\sqrt{16}}\cdot\sqrt{\sqrt{8}}=

Video Solution

Solution Steps

00:00 Solve the following problem
00:03 A 'regular' root raised to the second power
00:11 When there is a root of order (C) of root (B)
00:15 The result equals the root of the orders' product
00:22 Apply this formula to our exercise
00:29 When we have a product of 2 numbers (A and B) in a root of order (C)
00:33 The result equals their product (A times B) in a root of order (C)
00:36 Apply this formula to our exercise, and proceed to calculate the product
00:46 This is the solution

Step-by-Step Solution

To solve the problem 168 \sqrt{\sqrt{16}}\cdot\sqrt{\sqrt{8}} , we will follow these steps:

  • Step 1: Simplify each nested square root expression.
  • Step 2: Multiply the simplified expressions of the roots.

Step 1: Evaluate 16 \sqrt{\sqrt{16}} .
Since 16 can be expressed as 24 2^4 , we have: 16=(161/2)1/2=161/4=(24)1/4=24/4=21=2 \sqrt{\sqrt{16}} = \left(16^{1/2}\right)^{1/2} = 16^{1/4} = \left(2^4\right)^{1/4} = 2^{4/4} = 2^{1} = 2

Evaluate 8 \sqrt{\sqrt{8}} .
Since 8 can be expressed as 23 2^3 , we have: 8=(81/2)1/2=81/4=(23)1/4=23/4 \sqrt{\sqrt{8}} = \left(8^{1/2}\right)^{1/2} = 8^{1/4} = \left(2^3\right)^{1/4} = 2^{3/4}

Step 2: Multiply these simplified expressions together: 223/4=2123/4=21+3/4=27/4 2 \cdot 2^{3/4} = 2^{1} \cdot 2^{3/4} = 2^{1 + 3/4} = 2^{7/4}

Finally, converting back to radical form: 27/4=(27)1/4=274=1284 2^{7/4} = \left(2^7\right)^{1/4} = \sqrt[4]{2^7} = \sqrt[4]{128}

Thus, the solution to the problem is 1284 \sqrt[4]{128} , which corresponds to answer choice 3.

Answer

1284 \sqrt[4]{128}