Question

multiply. assume f is greater than or equal to zero, and write your answer in simplest form.\\(sqrt{2f^2} cdot sqrt{18f}\\)

Explanation:

Step1: Use product rule for radicals

The product rule for radicals states that \(\sqrt{a} \cdot \sqrt{b}=\sqrt{ab}\) (where \(a\geq0,b\geq0\)). So we can combine the two square roots:
\(\sqrt{2f^{2}}\cdot\sqrt{18f}=\sqrt{(2f^{2})\cdot(18f)}\)

Step2: Simplify the expression inside the square root

First, multiply the coefficients and the variables separately. For the coefficients: \(2\times18 = 36\). For the variables: \(f^{2}\times f=f^{2 + 1}=f^{3}\) (using the rule \(a^{m}\cdot a^{n}=a^{m + n}\)). So the expression inside the square root becomes \(36f^{3}\), and we have \(\sqrt{36f^{3}}\)

Step3: Simplify the square root

We can split \(\sqrt{36f^{3}}\) into \(\sqrt{36}\cdot\sqrt{f^{2}}\cdot\sqrt{f}\) (since \(f^{3}=f^{2}\cdot f\) and using the product rule for radicals again). We know that \(\sqrt{36} = 6\) and \(\sqrt{f^{2}}=f\) (because \(f\geq0\)). So putting it together: \(6\times f\times\sqrt{f}=6f\sqrt{f}\)

Answer:

\(6f\sqrt{f}\)