Prime Factorization of 441

Oct 13, 2025 by miodeya.com 19 views

Understanding the Problem

To simplify the given expression, we need to start by finding the prime factorization of both the numerator and the denominator. The prime factorization of a number is the expression of that number as the product of its prime factors.

Prime Factorization of 25

The prime factorization of 25 is 5^2. This is because 25 can be divided by 5, and when we divide 25 by 5, we get 5. Since 5 is a prime number, we cannot divide it further. Therefore, the prime factorization of 25 is 5^2.

Prime Factorization of 441

The prime factorization of 441 is 3^2 * 7^2. This is because 441 can be divided by 3, and when we divide 441 by 3, we get 147. We can further divide 147 by 3, and we get 49. Since 49 is a perfect square, we can take the square root of 49, and we get 7. Therefore, the prime factorization of 441 is 3^2 * 7^2.

Simplifying the Expression

Now that we have the prime factorization of both the numerator and the denominator, we can simplify the expression. We can rewrite the expression as $\sqrt{\frac{5^2}{3^2 * 7^2}}$ . Since the square root of a fraction is equal to the square root of the numerator divided by the square root of the denominator, we can rewrite the expression as $\frac{\sqrt{5^2}}{\sqrt{3^2 * 7^2}}$ .

Canceling Out Common Factors

We can simplify the expression further by canceling out common factors. Since both the numerator and the denominator have a factor of 5, we can cancel out the 5's. Similarly, since both the numerator and the denominator have a factor of 7, we can cancel out the 7's. Therefore, the expression simplifies to $\frac{\sqrt{5}}{\sqrt{3^2}}$ .

Simplifying the Square Root

We can simplify the square root further by taking the square root of the denominator. Since the square root of 3^2 is 3, we can rewrite the expression as $\frac{\sqrt{5}}{3}$ .

Conclusion

In conclusion, the expression $\sqrt{\frac{25}{441}}$ in simplest form is $\frac{\sqrt{5}}{3}$ .

Final Answer

The final answer is $\boxed{\frac{\sqrt{5}}{3}}$ .

Discussion

This problem requires us to find the prime factorization of both the numerator and the denominator, and then simplify the expression by canceling out common factors. The key concept here is the square root of a fraction, which is equal to the square root of the numerator divided by the square root of the denominator. By simplifying the expression step by step, we can arrive at the final answer.

Practice Problems

Simplify: $\sqrt{\frac{16}{81}}$
Simplify: $\sqrt{\frac{25}{100}}$
Simplify: $\sqrt{\frac{36}{49}}$

Real-World Applications

This problem has real-world applications in various fields such as engineering, physics, and computer science. For example, in engineering, we may need to simplify complex expressions involving square roots to arrive at a final answer. In physics, we may need to simplify expressions involving square roots to describe the motion of objects. In computer science, we may need to simplify expressions involving square roots to optimize algorithms.

Conclusion

In conclusion, simplifying expressions involving square roots is an important concept in mathematics that has real-world applications in various fields. By following the steps outlined in this problem, we can simplify complex expressions and arrive at a final answer.

Frequently Asked Questions

Q: What is the prime factorization of 25?

A: The prime factorization of 25 is 5^2. This is because 25 can be divided by 5, and when we divide 25 by 5, we get 5. Since 5 is a prime number, we cannot divide it further.

Q: What is the prime factorization of 441?

A: The prime factorization of 441 is 3^2 * 7^2. This is because 441 can be divided by 3, and when we divide 441 by 3, we get 147. We can further divide 147 by 3, and we get 49. Since 49 is a perfect square, we can take the square root of 49, and we get 7.

Q: How do we simplify the expression $\sqrt{\frac{25}{441}}$ ?

A: We can simplify the expression by finding the prime factorization of both the numerator and the denominator, and then canceling out common factors. Since both the numerator and the denominator have a factor of 5, we can cancel out the 5's. Similarly, since both the numerator and the denominator have a factor of 7, we can cancel out the 7's.

Q: What is the simplified form of the expression $\sqrt{\frac{25}{441}}$ ?

A: The simplified form of the expression $\sqrt{\frac{25}{441}}$ is $\frac{\sqrt{5}}{3}$ .

Q: Why do we need to simplify expressions involving square roots?

A: We need to simplify expressions involving square roots to arrive at a final answer. Simplifying expressions involving square roots can help us understand the underlying math and arrive at a more accurate answer.

Q: What are some real-world applications of simplifying expressions involving square roots?

A: Simplifying expressions involving square roots has real-world applications in various fields such as engineering, physics, and computer science. For example, in engineering, we may need to simplify complex expressions involving square roots to arrive at a final answer. In physics, we may need to simplify expressions involving square roots to describe the motion of objects. In computer science, we may need to simplify expressions involving square roots to optimize algorithms.

Q: How do we know if an expression involving a square root can be simplified?

A: We can determine if an expression involving a square root can be simplified by finding the prime factorization of both the numerator and the denominator, and then canceling out common factors.

Q: What is the final answer to the expression $\sqrt{\frac{25}{441}}$ ?

A: The final answer to the expression $\sqrt{\frac{25}{441}}$ is $\boxed{\frac{\sqrt{5}}{3}}$ .

Additional Resources

Practice Problems

Simplify: $\sqrt{\frac{16}{81}}$
Simplify: $\sqrt{\frac{25}{100}}$
Simplify: $\sqrt{\frac{36}{49}}$

Real-World Applications

Simplifying expressions involving square roots has real-world applications in various fields such as engineering, physics, and computer science.
In engineering, we may need to simplify complex expressions involving square roots to arrive at a final answer.
In physics, we may need to simplify expressions involving square roots to describe the motion of objects.
In computer science, we may need to simplify expressions involving square roots to optimize algorithms.