Question

Simplify 54−−√3
.

Answer 1

What is the irrational square root 63−−√
rewritten as the product of an integer and another irrational square root, in simplest form?(1 point)

Answer 2

To simplify the expression \( 54\sqrt{3} \), we can rewrite \( 54 \) as a product of its prime factors.

1. First, factor \( 54 \):
\[
54 = 2 \times 27 = 2 \times 3^3
\]

2. Now, substitute this factorization back into the expression:
\[
54\sqrt{3} = (2 \times 3^3)\sqrt{3}
\]

3. Distributing the square root, we get:
\[
= 2 \times 3^3 \times \sqrt{3} = 2 \times 3^3 \times 3^{1/2}
\]

4. Combine the powers of \( 3 \):
\[
= 2 \times 3^{3 + 1/2} = 2 \times 3^{3.5} = 2 \times 3^{7/2}
\]

5. Therefore, the simplified expression using the exponent:
\[
= 2 \sqrt{3^7} = 2 \sqrt{2187}
\]

6. Alternatively, rephrasing it in terms of simpler radicals, we can represent \( 3^{7/2} \) as:
\[
= 2 \times 3^3 \sqrt{3} = 2 \times 27\sqrt{3} = 54\sqrt{3}
\]

So, the simplified form is \( 54\sqrt{3} \).

Thus, the final simplified form remains:
\[
54\sqrt{3}.
\]

Answer 3

To simplify the expression \( \sqrt{63} \), we can start by factoring \( 63 \) into its prime factors:

1. **Factor \( 63 \)**:
\[
63 = 9 \times 7 = 3^2 \times 7
\]

2. **Rewrite the square root**:
\[
\sqrt{63} = \sqrt{9 \times 7}
\]

3. **Apply the property of square roots**:
\[
\sqrt{63} = \sqrt{9} \times \sqrt{7}
\]

4. **Calculate \( \sqrt{9} \)**:
\[
\sqrt{9} = 3
\]

5. **Combine the results**:
\[
\sqrt{63} = 3\sqrt{7}
\]

Thus, the expression \( \sqrt{63} \) rewritten as the product of an integer and another irrational square root in simplest form is:

\[
3\sqrt{7}
\]