Question
what is the irrational square root radical 72 rewritten as the product of an integer and another irrational square root
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Answered by
GPT-4o mini
To rewrite the irrational square root of \( 72 \) as the product of an integer and another irrational square root, we first factor \( 72 \) into its prime factors:
\[
72 = 2^3 \times 3^2
\]
Next, we can express \( \sqrt{72} \) as follows:
\[
\sqrt{72} = \sqrt{2^3 \times 3^2}
\]
We can separate this into two parts:
\[
\sqrt{72} = \sqrt{2^2 \times 2^1 \times 3^2}
\]
Using the property of square roots that states \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \), we can rewrite this as:
\[
\sqrt{72} = \sqrt{2^2} \times \sqrt{2^1} \times \sqrt{3^2}
\]
Calculating the square roots of the perfect squares:
\[
\sqrt{2^2} = 2 \quad \text{and} \quad \sqrt{3^2} = 3
\]
Thus, we have:
\[
\sqrt{72} = 2 \times 3 \times \sqrt{2} = 6 \sqrt{2}
\]
Therefore, the square root of \( 72 \) rewritten as the product of an integer and another irrational square root is:
\[
\sqrt{72} = 6 \sqrt{2}
\]
\[
72 = 2^3 \times 3^2
\]
Next, we can express \( \sqrt{72} \) as follows:
\[
\sqrt{72} = \sqrt{2^3 \times 3^2}
\]
We can separate this into two parts:
\[
\sqrt{72} = \sqrt{2^2 \times 2^1 \times 3^2}
\]
Using the property of square roots that states \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \), we can rewrite this as:
\[
\sqrt{72} = \sqrt{2^2} \times \sqrt{2^1} \times \sqrt{3^2}
\]
Calculating the square roots of the perfect squares:
\[
\sqrt{2^2} = 2 \quad \text{and} \quad \sqrt{3^2} = 3
\]
Thus, we have:
\[
\sqrt{72} = 2 \times 3 \times \sqrt{2} = 6 \sqrt{2}
\]
Therefore, the square root of \( 72 \) rewritten as the product of an integer and another irrational square root is:
\[
\sqrt{72} = 6 \sqrt{2}
\]
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