What is the irrational square root √63 rewritten as the product of an integer and another irrational square root, in simplest form?
To rewrite √63 as the product of an integer and another irrational square root, we need to find a square number that can be factored out from 63.
The prime factorization of 63 is 3 × 3 × 7. Since there are only single occurrences of each prime factor, we cannot factor out any square number from 63.
Therefore, √63 cannot be rewritten as the product of an integer and another irrational square root in simplest form.
To rewrite the irrational square root √63 as the product of an integer and another irrational square root in simplest form, we need to find the largest perfect square that divides evenly into 63.
The largest perfect square that divides evenly into 63 is 9, which is equal to 3^2. We can rewrite 63 as 9 * 7.
So, we can rewrite √63 as √(9 * 7).
Taking the square root of 9, we get 3. Therefore, we can rewrite √63 as 3 * √7.
Thus, the irrational square root √63 can be written as the product of an integer (3) and another irrational square root (√7), in simplest form.
To rewrite the irrational square root √63 as the product of an integer and another irrational square root, we first need to find the factors of 63. The factors of 63 are 1, 3, 7, 9, 21, and 63.
Next, we look for a perfect square among these factors. A perfect square is a number whose square root is an integer. From the factors of 63, we can see that 9 is a perfect square because √9 = 3.
Now, we can rewrite √63 as the product of √9 and another irrational square root. Simplifying √9 gives us 3. Hence, we can write √63 as 3√7 in simplest form.
Therefore, the irrational square root √63 can be rewritten as 3√7.