Question

which expression is equivalent to \\(\sqrt{\frac{4j^4}{9k^8}}\\)?
\\(\circ\\) \\(\frac{2}{3}j^2k^4\\)
\\(\circ\\) \\(\frac{2j^2}{3k^4}\\)
\\(\circ\\) \\(\frac{2}{3}j^2k^4\\)
\\(\circ\\) \\(\frac{2j^2}{3k^4}\\)

Explanation:

Step1: Simplify the square root of the fraction

We know that $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$, so $\sqrt{\frac{4j^{4}}{9k^{8}}}=\frac{\sqrt{4j^{4}}}{\sqrt{9k^{8}}}$.

Step2: Simplify the square roots of the numerator and denominator

For the numerator, $\sqrt{4j^{4}}=\sqrt{4}\times\sqrt{j^{4}} = 2j^{2}$ (since $\sqrt{4} = 2$ and $\sqrt{j^{4}}=j^{2}$ as $(j^{2})^{2}=j^{4}$).
For the denominator, $\sqrt{9k^{8}}=\sqrt{9}\times\sqrt{k^{8}}=3k^{4}$ (since $\sqrt{9} = 3$ and $\sqrt{k^{8}}=k^{4}$ as $(k^{4})^{2}=k^{8}$).

Step3: Combine the simplified numerator and denominator

So $\frac{\sqrt{4j^{4}}}{\sqrt{9k^{8}}}=\frac{2j^{2}}{3k^{4}}$.

Answer:

$\frac{2j^{2}}{3k^{4}}$ (assuming the last option is $\frac{2j^{2}}{3k^{4}}$ as per the formatting, the correct option is the last one with $\frac{2j^{2}}{3k^{4}}$)