Question

simplify.
\sqrt{18x^2} \sqrt{2x}
assume that the variable represents a positive real number.

Explanation:

Step1: Multiply the radicands

Using the property $\sqrt{a}\sqrt{b}=\sqrt{ab}$, we have $\sqrt{18x^{2}}\sqrt{2x}=\sqrt{(18x^{2})(2x)}$.
Calculating the product inside the square root: $(18x^{2})(2x)=18\times2\times x^{2}\times x = 36x^{3}$.
So now we have $\sqrt{36x^{3}}$.

Step2: Simplify the square root

We can rewrite $36x^{3}$ as $36x^{2}\cdot x$.
Then $\sqrt{36x^{3}}=\sqrt{36x^{2}\cdot x}$.
Using the property $\sqrt{ab}=\sqrt{a}\sqrt{b}$ (where $a = 36x^{2}$ and $b = x$), we get $\sqrt{36x^{2}}\cdot\sqrt{x}$.
Since $\sqrt{36x^{2}} = 6x$ (because $x$ is a positive real number, so $\sqrt{x^{2}}=x$), we have $6x\sqrt{x}$.

Answer:

$6x\sqrt{x}$