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The expression to simplify is:

\[
\sqrt{50v^{14}}
\]

### Step-by-step solution:

1. **Break down the square root**:
   \[
   \sqrt{50v^{14}} = \sqrt{50} \cdot \sqrt{v^{14}}
   \]

2. **Simplify \(\sqrt{50}\)**:
   \[
   \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}
   \]

3. **Simplify \(\sqrt{v^{14}}\)**:
   \[
   \sqrt{v^{14}} = v^{14/2} = v^7
   \]

4. **Final simplification**:
   Combine the terms:
   \[
   \sqrt{50v^{14}} = 5\sqrt{2} \cdot v^7 = 5v^7\sqrt{2}
   \]

So, the simplified expression is:

\[
5v^7\sqrt{2}
\]

Would you like more details or have any questions?

Here are five related questions to extend your understanding:
1. How would you simplify \(\sqrt{18v^{12}}\)?
2. Can you explain why \(\sqrt{v^{14}} = v^7\)?
3. What happens if the exponent of \(v\) is odd inside a square root?
4. How would the process change if it were a cube root instead of a square root?
5. Can you break down \(\sqrt{72v^8}\) in a similar way?

**Tip**: When simplifying square roots, factorize numbers and variables completely to handle roots efficiently.

Simplify the expression \(\sqrt{50v^{14}}\), assuming that the variable \(v\) represents a positive real number.

Learn how to simplify the expression \(\sqrt{50v^{14}}\), using algebraic techniques such as breaking down square roots and applying laws of exponents. This solution is suitable for students in Grades 9-10.

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