Answer :
To simplify \(\sqrt{484}\), we'll follow the given steps:
1. Write the prime factorization of the radicand:
[tex]\[ \sqrt{484} = \sqrt{2 \cdot 2 \cdot 11 \cdot 11} \][/tex]
2. Apply the product property of square roots: Write the radicand as a product, forming as many perfect squares as possible.
[tex]\[ \sqrt{2 \cdot 2 \cdot 11 \cdot 11} = \sqrt{2^2 \cdot 11^2} \][/tex]
3. Simplify: Take the square root of each perfect square term.
[tex]\[ \sqrt{2^2 \cdot 11^2} = \sqrt{2^2} \cdot \sqrt{11^2} \][/tex]
Now, we simplify the square roots of the perfect squares:
[tex]\[ \sqrt{2^2} = 2 \quad \text{and} \quad \sqrt{11^2} = 11 \][/tex]
4. Combine the results:
[tex]\[ \sqrt{2^2} \cdot \sqrt{11^2} = 2 \cdot 11 \][/tex]
Therefore,
[tex]\[ \sqrt{484} = 2 \cdot 11 = 22 \][/tex]
So, the simplified form of [tex]\(\sqrt{484}\)[/tex] is [tex]\(22\)[/tex].
1. Write the prime factorization of the radicand:
[tex]\[ \sqrt{484} = \sqrt{2 \cdot 2 \cdot 11 \cdot 11} \][/tex]
2. Apply the product property of square roots: Write the radicand as a product, forming as many perfect squares as possible.
[tex]\[ \sqrt{2 \cdot 2 \cdot 11 \cdot 11} = \sqrt{2^2 \cdot 11^2} \][/tex]
3. Simplify: Take the square root of each perfect square term.
[tex]\[ \sqrt{2^2 \cdot 11^2} = \sqrt{2^2} \cdot \sqrt{11^2} \][/tex]
Now, we simplify the square roots of the perfect squares:
[tex]\[ \sqrt{2^2} = 2 \quad \text{and} \quad \sqrt{11^2} = 11 \][/tex]
4. Combine the results:
[tex]\[ \sqrt{2^2} \cdot \sqrt{11^2} = 2 \cdot 11 \][/tex]
Therefore,
[tex]\[ \sqrt{484} = 2 \cdot 11 = 22 \][/tex]
So, the simplified form of [tex]\(\sqrt{484}\)[/tex] is [tex]\(22\)[/tex].
