What is the product of 6, square root of, 12, end square root6

Question

12
​
and 2, square root of, 14, end square root2
14
​
in simplest radical fo

Answer 1

To find the product of \( 6 \sqrt{12} \) and \( 2 \sqrt{14} \), we can follow these steps:

1. First, multiply the coefficients (the whole numbers) together.
2. Next, multiply the square root parts together.
3. Finally, simplify the expression if possible.

Let's start by calculating the product:

\[ 6 \sqrt{12} \times 2 \sqrt{14} \]

1. Multiply the coefficients: \[ 6 \times 2 = 12 \]

2. Multiply the square roots: \[ \sqrt{12} \times \sqrt{14} = \sqrt{12 \times 14} \] Calculating \( 12 \times 14 \): \[ 12 \times 14 = 168 \]

So, we have: \[ \sqrt{12} \times \sqrt{14} = \sqrt{168} \]

3. Now we combine the results: \[ 12 \sqrt{168} \]

4. Next, we simplify \( \sqrt{168} \). We can factor \( 168 \): \[ 168 = 2^3 \times 3 \times 7 \] Taking the square root: \[ \sqrt{168} = \sqrt{(2^2) \cdot 2 \cdot 3 \cdot 7} = 2 \sqrt{42} \]

5. Substitute back into the expression: \[ 12 \sqrt{168} = 12 \times 2 \sqrt{42} = 24 \sqrt{42} \]

Thus, the product \( 6 \sqrt{12} \times 2 \sqrt{14} \) in simplest radical form is:

\[ \boxed{24 \sqrt{42}} \]