Question

sum of 3 radical 16 and 4 radical 6 in simplest form, also, determine whether the result is rational or irrational and explain your answer

Answer 1

To sum \(3\sqrt{16}\) and \(4\sqrt{6}\), we first simplify each term:

1. For \(3\sqrt{16}\):
\[
\sqrt{16} = 4,
\]
so:
\[
3\sqrt{16} = 3 \times 4 = 12.
\]

2. For \(4\sqrt{6}\), it remains as \(4\sqrt{6}\) since \(6\) is not a perfect square and cannot be simplified further.

Now we can add the two results together:
\[
12 + 4\sqrt{6}.
\]

This expression, \(12 + 4\sqrt{6}\), is the sum of a rational number (12) and an irrational number (\(4\sqrt{6}\)).

Now, we need to determine if the overall result is rational or irrational.

A rational number is any number that can be expressed as the quotient of two integers. An irrational number cannot be expressed this way and has a non-repeating, non-terminating decimal expansion.

Since we have a rational number (12) and an irrational number (\(4\sqrt{6}\)), the sum of a rational and an irrational number is always irrational.

Thus, the final answer is:
\[
\text{The sum is } 12 + 4\sqrt{6}, \text{ which is irrational.}
\]