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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]
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].
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