Bill and Jill were asked to simplify √72 Bill got an answer of 3√8 and Jill got an answer of 6√2. who is correct? Explain below why they are correct and/or why the other person is incorrect.

Answer 1

To simplify \( \sqrt{72} \), we can break it down into its prime factors:

1. First, we find the prime factorization of 72: \[ 72 = 8 \times 9 = 2^3 \times 3^2 \]

2. Now, we can simplify \( \sqrt{72} \) using these factors: \[ \sqrt{72} = \sqrt{2^3 \times 3^2} \]

3. Using the properties of square roots, we can separate the square roots of the factors: \[ \sqrt{72} = \sqrt{2^3} \times \sqrt{3^2} \]

4. Simplifying each term, we get:

   • \( \sqrt{2^3} = \sqrt{(2^2 \cdot 2)} = 2\sqrt{2} \)
   • \( \sqrt{3^2} = 3 \)

5. Thus, we write: \[ \sqrt{72} = 2\sqrt{2} \times 3 = 6\sqrt{2} \]

Now, let's evaluate Bill's answer of \( 3\sqrt{8} \):

1. We can simplify \( \sqrt{8} \): \[ \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \]

2. Substituting this back into Bill's expression: \[ 3\sqrt{8} = 3 \times 2\sqrt{2} = 6\sqrt{2} \]

Conclusion:

Both Bill and Jill arrive at equivalent answers when simplifying \( \sqrt{72} \):

• Jill's answer is \( 6\sqrt{2} \).
• Bill's answer of \( 3\sqrt{8} = 6\sqrt{2} \).

Thus, both Bill and Jill are correct; their answers are both valid and equivalent.

Answer 2

choose one right answer and explain why

Answer 3

The correct simplification of \( \sqrt{72} \) is Jill's answer: \( 6\sqrt{2} \).

Explanation:

1. Simplifying \( \sqrt{72} \):

   • First, we find the prime factorization of 72: \[ 72 = 2^3 \times 3^2 \]
   • To simplify \( \sqrt{72} \), we can separate the square root of the factors: \[ \sqrt{72} = \sqrt{(2^3) \times (3^2)} = \sqrt{2^3} \times \sqrt{3^2} \]
   • Simplifying each part:
     • \( \sqrt{3^2} = 3 \)
     • \( \sqrt{2^3} = \sqrt{(2^2) \cdot 2} = 2\sqrt{2} \)

2. Combining the results: \[ \sqrt{72} = 2\sqrt{2} \times 3 = 6\sqrt{2} \]

3. Bill's answer:

   • Bill's answer was \( 3\sqrt{8} \).
   • We can simplify \( \sqrt{8} \): \[ \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \]
   • So, Bill's expression becomes: \[ 3\sqrt{8} = 3 \times 2\sqrt{2} = 6\sqrt{2} \]

Conclusion:

Though both answers ultimately simplify to \( 6\sqrt{2} \), Jill's answer is presented in the simplest and most direct form. Therefore, it is more straightforward to say she provided the correct answer. Bill's answer, while equivalent, takes an extra step that isn't necessary in the simplification process.

Thus, the final answer is: Jill's \( 6\sqrt{2} \) is the preferred and directly simplified answer.

Answer 4

why is jill right

Answer 5

Jill is right because her answer, \( 6\sqrt{2} \), is the correct and simplest form of the expression \( \sqrt{72} \).

Explanation:

1. Correct Simplification of \( \sqrt{72} \):

   • We can simplify \( \sqrt{72} \) using its prime factorization.
   • The factorization of 72 is \( 72 = 2^3 \times 3^2 \).
   • We apply the square root to this factorization: \[ \sqrt{72} = \sqrt{2^3 \times 3^2} = \sqrt{2^2 \times 2 \times 3^2} \]

2. Breaking It Down:

   • We can separate this into: \[ \sqrt{72} = \sqrt{2^2} \times \sqrt{2} \times \sqrt{3^2} \]
   • Simplifying each part:
     • \( \sqrt{2^2} = 2 \)
     • \( \sqrt{3^2} = 3 \)
   • So we have: \[ \sqrt{72} = 2 \times 3 \times \sqrt{2} = 6\sqrt{2} \]

3. Bill's Answer:

   • Bill's answer was \( 3\sqrt{8} \).
   • While \( \sqrt{8} \) can be simplified to \( 2\sqrt{2} \), his expression took an unnecessary roundabout way to arrive at the same numerical answer.
   • Simplifying as follows: \[ \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \]
   • So Bill’s result becomes: \[ 3\sqrt{8} = 3 \cdot 2\sqrt{2} = 6\sqrt{2} \]
   • While both Jill's and Bill's final answers are essentially the same, Jill provided the direct simplification without alternate steps.

Conclusion:

Jill is right in the sense that her answer \( 6\sqrt{2} \) is the cleanest and most straightforward result from simplifying \( \sqrt{72} \). While both answers are equivalent, Jill's answer strips away any extraneous steps by directly arriving at the simplest form.