What is the irrational square root of 63 rewritten as the product of an integer and another irrational square root, in simplest form?
To rewrite the irrational square root of 63 as the product of an integer and another irrational square root, we can factorize 63 as 9 × 7. Since 9 is a perfect square, we can simplify the square root of 63 as the square root of 9 times the square root of 7:
√63 = √(9 × 7) = √9 × √7 = 3√7
Therefore, the irrational square root of 63 can be rewritten as 3 times the irrational square root of 7, in simplest form.