Use the product rule to simplify sqrt(3x) * sqrt(21x), assuming the variables represent nonnegative real numbers.

Step-by-step solution

1. 1

   Combine under a single radical. The product rule $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ is valid for $a, b \ge 0$, which the nonnegativity assumption guarantees:

   $\sqrt{3x}\cdot\sqrt{21x} = \sqrt{(3x)(21x)}$

2. 2

   Multiply the radicand.

   $(3x)(21x) = 63x^2 \quad\Longrightarrow\quad \sqrt{63x^2}$

3. 3

   Split off the perfect-square factors. Factor $63 = 9 \times 7$, where $9$ is a perfect square and $7$ is square-free:

   $\sqrt{63x^2} = \sqrt{9}\cdot\sqrt{7}\cdot\sqrt{x^2} = 3\sqrt{7}\cdot\sqrt{x^2}$

4. 4

   Simplify the variable radical. In general $\sqrt{x^2} = |x|$, but the problem states $x \ge 0$, so the absolute value is unnecessary and $\sqrt{x^2} = x$. Hence

   $\sqrt{3x}\cdot\sqrt{21x} = 3x\sqrt{7}$

5. 5

   Check numerically at x = 3. The original is $\sqrt{9}\cdot\sqrt{63} = 3 \times 7.937254 = 23.811762$. The simplified form gives $9\sqrt7 = 9 \times 2.645751 = 23.811762$. They agree to six decimals.