What is the irrational square root sqrt(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 need to find a perfect square that is a factor of 63.
The factors of 63 are 1, 3, 7, 9, 21, and 63. Among these, 9 is a perfect square.
Therefore, we can rewrite sqrt(63) as sqrt(9 * 7).
Using the properties of square roots, this can be simplified as sqrt(9) * sqrt(7).
Since the square root of 9 is equal to 3, the simplified form is 3 * sqrt(7).
To rewrite the irrational square root sqrt(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. We can rewrite 63 as 9 * 7.
Now, let's rewrite sqrt(63) using this information:
sqrt(63) = sqrt(9 * 7)
Since sqrt(9) = 3, we can remove the perfect square root of 9 and rewrite sqrt(63) as:
sqrt(63) = sqrt(9) * sqrt(7)
Simplifying further, we have:
sqrt(63) = 3 * sqrt(7)
Therefore, the irrational square root sqrt(63) can be rewritten as the product of the integer 3 and the irrational square root sqrt(7), in simplest form.
To rewrite the irrational square root √(63) as the product of an integer and another irrational square root, we need to find a perfect square that divides 63.
Let's break down 63 into its prime factors: 63 = 3 * 3 * 7.
Since there are no perfect squares in the factors of 63, we cannot directly simplify the square root. However, we can express it as the product of the square root of the largest perfect square we can extract from 63.
In this case, the largest perfect square that divides 63 is 9, which is 3 * 3. Therefore, we can rewrite the square root of 63 as follows:
√(63) = √(9 * 7) = √9 * √7
Simplifying further:
√9 is a perfect square and equals 3. Thus, we have:
√(63) = 3√7
Therefore, the irrational square root √(63) can be rewritten as 3√7, where 3 is an integer and √7 is another irrational square root.