What is the product of 9, square root of, 3, end square root9

Question

3
​
and 5, square root of, 21, end square root5
21
​
in simplest radical form?

Answer 1

To find the product of \( 9\sqrt{3} \) and \( 5\sqrt{21} \), we can follow these steps:

1. Multiply the coefficients and the radicals separately:

   \[ (9\sqrt{3}) \times (5\sqrt{21}) = 9 \times 5 \times \sqrt{3} \times \sqrt{21} \]

2. Calculate the coefficient:

   \[ 9 \times 5 = 45 \]

3. Multiply the radicals:

   \[ \sqrt{3} \times \sqrt{21} = \sqrt{3 \times 21} = \sqrt{63} \]

4. Combine the results:

   \[ 45 \times \sqrt{63} \]

5. Simplify \( \sqrt{63} \):

   The number \( 63 \) can be factored as:

   \[ 63 = 9 \times 7 = 3^2 \times 7 \]

   Therefore, we have:

   \[ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} = 3\sqrt{7} \]

6. Substitute back to get the final result:

   \[ 45\sqrt{63} = 45 \times 3\sqrt{7} = 135\sqrt{7} \]

Thus, the product of \( 9\sqrt{3} \) and \( 5\sqrt{21} \) in simplest radical form is

\[ \boxed{135\sqrt{7}} \]