Rewrite the irrational square root √48 as the product of an interrogator and another rational square root
To rewrite √48 as the product of an interrogator and another rational square root, we can first simplify √48.
√48 = √(4 × 12)
= √4 × √12
= 2 × √12
Now, let's simplify √12 further.
√12 = √(4 × 3)
= √4 × √3
= 2 × √3
Therefore, √48 = 2 × √3
So, √48 can be written as the product of an interrogator (2) and another rational square root (√3).
To rewrite the irrational square root √48 as the product of an interrogator and another rational square root, we can begin by finding the prime factorization of 48.
Prime factorization of 48:
48 = 2^4 * 3
Now, let's rearrange the terms under the square root:
√48 = √(2^4 * 3)
Since the square root is distributive over multiplication, we can split it up:
√48 = √(2^4) * √3
Simplifying each square root gives:
√48 = 2^2 * √3
Finally, we can express 2^2 as 4, so the simplified expression is:
√48 = 4√3
Therefore, the irrational square root √48 can be rewritten as the product of an interrogator (4) and the rational square root (√3).
To rewrite the irrational square root √48 as the product of an interrogator and another rational square root, let's first simplify the square root of 48.
We can start by finding the prime factorization of 48:
48 = 2 * 2 * 2 * 2 * 3 = (2^4) * 3
Next, let's identify any perfect square factors that we can take out from under the square root sign. In this case, we can take out 2^2 = 4 because it is a perfect square.
Therefore, we have:
√48 = √(4 * (2^2) * 3) = √(4) * √(2^2) * √3 = 2 * 2 * √3 = 4√3
So, rewriting the original irrational square root √48 as the product of an interrogator (4) and another rational square root (√3), we have:
√48 = 4√3