Find $\frac{3\sqrt{4a^{3}}}{5\sqrt{3a^{4}}}$

Question

$$\frac{3\sqrt{4a^{3}}}{5\sqrt{3a^{4}}}$$

I got this question on my math homework, but I'm not sure if the answer I got is the correct one. I'll link a picture of the work I did on it, along with the answer I got below.

Answer 1

Note that $a>0$,

\begin{align*} \frac{3\sqrt{4a^3}}{5\sqrt{3a^4}} &= \frac{3\times 2a\sqrt{a}}{5\times a^2\sqrt{3}} \\ &= \frac{2\sqrt{3} \times \sqrt{a}}{5a} \\ &= \frac{2\sqrt{3a}}{5a} \end{align*}

If no needs to rationalize the denominator, I prefer writing as $\fbox{$\dfrac{2\sqrt{3}}{5\sqrt{a}}$}$.

Answer 2

$$ \frac{3\sqrt{4a^3}}{5\sqrt{3a^4}}=\frac{3\sqrt{4}\sqrt{a^2}\sqrt{a}}{5\sqrt{3}\sqrt{(a^2)^2}}= \frac{3\cdot2\cdot a\cdot\sqrt{a}}{5\sqrt{3}\cdot a^2} =\frac{\sqrt{3}\sqrt{3}\cdot2\cdot a\cdot\sqrt{a}}{5\sqrt{3}\cdot a\cdot\sqrt{a}\cdot\sqrt{a}} =\frac{2\sqrt{3}}{5\sqrt{a}} $$

Answer 3

After your first step, you should have $\sqrt{a^8} = a^4$ on the bottom instead of $a^8$. You should end up with the answer $$ \frac{2\sqrt {3a} }{5 a} $$