Simplify the Nested Square Root: √(√(16x²))

Q: Why can't I just combine the square roots into one?

Nested square roots don't combine like multiplication! $ \sqrt{\sqrt{16x^2}} $ is not the same as $ \sqrt{16x^2} $. You must work from the inside out, one root at a time.

Q: What if x is negative?

For $ \sqrt{x^2} $ to be real, we typically assume x ≥ 0 in this context. When x is negative, $ \sqrt{x^2} = |x| = -x $, which changes our final answer.

Q: How do I know when to stop simplifying?

Stop when you have no perfect squares under any remaining radicals. In our case, $ 2\sqrt{x} $ is fully simplified because x alone has no perfect square factors we can extract.

Q: Can I use fractional exponents instead?

Yes! $ \sqrt{\sqrt{16x^2}} = (16x^2)^{1/4} = 16^{1/4} \cdot x^{1/2} = 2\sqrt{x} $. Both methods give the same answer, but working step-by-step with radicals often prevents errors.

Q: Why is the answer not 2x?

If the answer were 2x, then $ (2x)^2 = 4x^2 $, not 4x. Since our inner expression simplifies to 4x, we need $ 2\sqrt{x} $ because $ (2\sqrt{x})^2 = 4x $ ✓

Nested Square Roots with Variable Expressions

Complete the following exercise:

$\sqrt{\sqrt{16\cdot x^2}}=$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve the following problem

00:05 When we have a root of the multiplication (A times B)

00:09 We can write it as a multiplication of the root of each term

00:12 We'll apply this formula to our exercise and proceed to break down the root

00:18 Break down 16 to 4 squared

00:22 The root of any number (A) squared cancels the square

00:30 Apply this formula to our exercise and cancel the squares

00:49 Break down 4 to 2 squared

00:53 The root cancels the square

00:56 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:

$\sqrt{\sqrt{16\cdot x^2}}=$

Step-by-step solution

To solve the expression $\sqrt{\sqrt{16 \cdot x^2}}$ , follow these steps:

Step 1: Simplify the innermost root $\sqrt{16 \cdot x^2}$ .
Here, apply $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ . Thus, $\sqrt{16 \cdot x^2} = \sqrt{16} \cdot \sqrt{x^2}$ .
Step 2: Calculate each component:
- $\sqrt{16} = 4$ because $16^{1/2} = 4$ .
- $\sqrt{x^2} = x$ assuming $x$ is non-negative.
Step 3: Combine results from Step 2: $\sqrt{16} \cdot \sqrt{x^2} = 4 \cdot x = 4x$ .
Step 4: Simplify the outer square root: $\sqrt{4x}$ .
Applying $\sqrt{4x} = \sqrt{4} \cdot \sqrt{x}$ , we have $\sqrt{4} = 2$ .
Thus, $\sqrt{4x} = 2 \cdot \sqrt{x} = 2\sqrt{x}$ .

Therefore, the simplified form of $\sqrt{\sqrt{16 \cdot x^2}}$ is $2\sqrt{x}$ . This corresponds to choice 1.

Final Answer

$2\sqrt{x}$

Key Points to Remember

Essential concepts to master this topic

Property: Simplify from inside out when square roots are nested
Technique: $\sqrt{16x^2} = 4x$ , then $\sqrt{4x} = 2\sqrt{x}$
Check: Verify that $(2\sqrt{x})^2 = 4x$ matches inner expression ✓

Common Mistakes

Avoid these frequent errors

Combining exponents incorrectly in nested roots
Don't simplify $\sqrt{\sqrt{16x^2}}$ as $16^{1/4} \cdot x^{1/2} = 2x^{1/2}$ directly! This skips the crucial step of simplifying the inner root first. Always work from the innermost root outward: first find $\sqrt{16x^2} = 4x$ , then $\sqrt{4x} = 2\sqrt{x}$ .

Practice Quiz

Test your knowledge with interactive questions

Solve the following exercise:

$\sqrt[5]{\sqrt[3]{5}}=$

FAQ

Everything you need to know about this question

Why can't I just combine the square roots into one?

Nested square roots don't combine like multiplication! $\sqrt{\sqrt{16x^2}}$ is not the same as $\sqrt{16x^2}$ . You must work from the inside out, one root at a time.

What if x is negative?

For $\sqrt{x^2}$ to be real, we typically assume x ≥ 0 in this context. When x is negative, $\sqrt{x^2} = |x| = -x$ , which changes our final answer.

How do I know when to stop simplifying?

Stop when you have no perfect squares under any remaining radicals. In our case, $2\sqrt{x}$ is fully simplified because x alone has no perfect square factors we can extract.

Can I use fractional exponents instead?

Yes! $\sqrt{\sqrt{16x^2}} = (16x^2)^{1/4} = 16^{1/4} \cdot x^{1/2} = 2\sqrt{x}$ . Both methods give the same answer, but working step-by-step with radicals often prevents errors.

Why is the answer not 2x?

If the answer were 2x, then $(2x)^2 = 4x^2$ , not 4x. Since our inner expression simplifies to 4x, we need $2\sqrt{x}$ because $(2\sqrt{x})^2 = 4x$ ✓

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