Simplifying Perfect Square Roots

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]

[tex]\boxed{22}[/tex]



Answer :

To simplify the square root of \( \sqrt{484} \), we will follow the steps you outlined. Here's a detailed step-by-step solution:

1. Write the prime factorization of the radicand:

We start with the number 484 and find its prime factors. The prime factorization of 484 is:
[tex]\[ 484 = 2 \times 2 \times 11 \times 11 \][/tex]
Therefore:
[tex]\[ \sqrt{484} = \sqrt{2 \times 2 \times 11 \times 11} \][/tex]

2. Apply the product property of square roots:

The product property of square roots allows us to split the square root of a product into the product of square roots. Using this property, we can rewrite the radicand as a product of squares:
[tex]\[ \sqrt{2 \times 2 \times 11 \times 11} = \sqrt{2^2} \times \sqrt{11^2} \][/tex]

3. Simplify:

Now, we simplify the square roots of the perfect squares:
[tex]\[ \sqrt{2^2} = 2 \][/tex]
[tex]\[ \sqrt{11^2} = 11 \][/tex]
Therefore:
[tex]\[ \sqrt{2^2} \times \sqrt{11^2} = 2 \times 11 = 22 \][/tex]

Putting it all together, we have:
[tex]\[ \sqrt{484} = 22 \][/tex]

Thus, the simplified form of [tex]\( \sqrt{484} \)[/tex] is [tex]\( 22 \)[/tex].