Simplify 54−−√3

Question

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Answer 1

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 2

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 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:

\[ 3\sqrt{7} \]