Simplify: [tex]\sqrt{484}[/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 square roots as possible.
[tex]\sqrt{2 \cdot 2 \cdot 11 \cdot 11} = \sqrt{2^2} \cdot \sqrt{11^2}[/tex]

3. Simplify.
[tex]\sqrt{484} = 2 \cdot 11 = 22[/tex]



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].