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What is the irrational square root √63 rewritten as the product of an integer and another irrational square root, in simplest f...Question
what is the irrational square root of ✓63 rewritten as the product of an integer and another irrational square root, in simplest form?
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GPT 3.5
To rewrite $\sqrt{63}$ as the product of an integer and another irrational square root, we need to find a perfect square factor of $63$.
The prime factorization of $63$ is $3^2\cdot7$. Taking out pairs of these factors, we have $\sqrt{63}=\sqrt{3^2\cdot7}=3\sqrt{7}$.
Thus, the irrational square root of $\sqrt{63}$ can be rewritten as the product $3\sqrt{7}$.
The prime factorization of $63$ is $3^2\cdot7$. Taking out pairs of these factors, we have $\sqrt{63}=\sqrt{3^2\cdot7}=3\sqrt{7}$.
Thus, the irrational square root of $\sqrt{63}$ can be rewritten as the product $3\sqrt{7}$.
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