Question
Find the sum of 3, square root of, 5, end square root3
5
and 4, square root of, 8, end square root4
8
in simplest form. Also, determine whether the result is rational or irrational and explain your answer.
Answer
Attempt 1 out of 5
Result:
The result is
because it
be written as the ratio of two integers and its decimal expansion
terminate or repeat.
5
and 4, square root of, 8, end square root4
8
in simplest form. Also, determine whether the result is rational or irrational and explain your answer.
Answer
Attempt 1 out of 5
Result:
The result is
because it
be written as the ratio of two integers and its decimal expansion
terminate or repeat.
Answers
Answered by
GPT-4o mini
To find the sum of \( 3\sqrt{5} \) and \( 4\sqrt{8} \), let's first simplify \( 4\sqrt{8} \).
Since \( \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2} \), we can rewrite \( 4\sqrt{8} \) as:
\[
4\sqrt{8} = 4(2\sqrt{2}) = 8\sqrt{2}
\]
Next, we sum \( 3\sqrt{5} \) and \( 8\sqrt{2} \):
\[
3\sqrt{5} + 8\sqrt{2}
\]
Now we cannot combine \( 3\sqrt{5} \) and \( 8\sqrt{2} \) further because they are not like terms.
Thus, the sum in simplest form is:
\[
3\sqrt{5} + 8\sqrt{2}
\]
Now, to determine whether this result is rational or irrational, we need to analyze each term:
- The number \( \sqrt{5} \) is an irrational number.
- The number \( \sqrt{2} \) is also an irrational number.
When we multiply a rational number (like \( 3 \) or \( 8 \)) by an irrational number, the result is still irrational. Therefore:
- \( 3\sqrt{5} \) is irrational.
- \( 8\sqrt{2} \) is also irrational.
The sum of two irrational numbers can be rational or irrational. In this case, \( 3\sqrt{5} + 8\sqrt{2} \) cannot be expressed as a ratio of two integers, which implies that the sum is irrational.
### Final answer:
The result is \( 3\sqrt{5} + 8\sqrt{2} \) because it cannot be written as the ratio of two integers and its decimal expansion does not terminate or repeat. Thus, the result is **irrational**.
Since \( \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2} \), we can rewrite \( 4\sqrt{8} \) as:
\[
4\sqrt{8} = 4(2\sqrt{2}) = 8\sqrt{2}
\]
Next, we sum \( 3\sqrt{5} \) and \( 8\sqrt{2} \):
\[
3\sqrt{5} + 8\sqrt{2}
\]
Now we cannot combine \( 3\sqrt{5} \) and \( 8\sqrt{2} \) further because they are not like terms.
Thus, the sum in simplest form is:
\[
3\sqrt{5} + 8\sqrt{2}
\]
Now, to determine whether this result is rational or irrational, we need to analyze each term:
- The number \( \sqrt{5} \) is an irrational number.
- The number \( \sqrt{2} \) is also an irrational number.
When we multiply a rational number (like \( 3 \) or \( 8 \)) by an irrational number, the result is still irrational. Therefore:
- \( 3\sqrt{5} \) is irrational.
- \( 8\sqrt{2} \) is also irrational.
The sum of two irrational numbers can be rational or irrational. In this case, \( 3\sqrt{5} + 8\sqrt{2} \) cannot be expressed as a ratio of two integers, which implies that the sum is irrational.
### Final answer:
The result is \( 3\sqrt{5} + 8\sqrt{2} \) because it cannot be written as the ratio of two integers and its decimal expansion does not terminate or repeat. Thus, the result is **irrational**.