Hello everyone! I would like it very much if you all checked my work. Thank you.

My directions are: Simplify. Assume that no radicands were formed by raising a negative quantities to even powers.

My problem is: $\frac{\sqrt{128a^2b^4}}{\sqrt{16ab}}$

I dwindled it down to: $\frac{\sqrt{64\cdot 2}ab^2}{4\sqrt{ab}}\;=\;\frac{8\sqrt{2}ab^2}{4\sqr t{ab}}\;=\;\frac{2\sqrt{2}ab^2}{\sqrt{ab}}$

So now I need to rationalize the denominator right?

So I did that and got: $\boxed{\frac{2ab^2\sqrt{2ab}}{ab}}$

Does this look good, my math community?

2. Originally Posted by Jonboy
Hello everyone! I would like it very much if you all checked my work. Thank you.

My directions are: Simplify. Assume that no radicands were formed by raising a negative quantities to even powers.

My problem is: $\frac{\sqrt{128a^2b^4}}{\sqrt{16ab}}$

I dwindled it down to: $\frac{\sqrt{64\cdot 2}ab^2}{4\sqrt{ab}}\;=\;\frac{8\sqrt{2}ab^2}{4\sqr t{ab}}\;=\;\frac{2\sqrt{2}ab^2}{\sqrt{ab}}$

So now I need to rationalize the denominator right?

So I did that and got: $\boxed{\frac{2ab^2\sqrt{2ab}}{ab}}$

Does this look good, my math community?
i didn't check your working in detail, but it is obvious that your final answer is not simplified. you have ab in the denominator when we can cancel them with the ab^2 in the numerator.

try this approach:

$\frac {\sqrt {128a^2b^4}}{\sqrt {16ab}} = \frac {128^{1/2}\left( a^2 \right)^{1/2}\left( b^4 \right)^{1/2}}{16^{1/2}a^{1/2}b^{1/2}} = \frac {128^{1/2}}{16^{1/2}} \cdot \frac {\left( a^2 \right)^{1/2}}{a^{1/2}} \cdot \frac {\left( b^4 \right)^{1/2}}{b^{1/2}}$

recall that $\left( x^a \right)^b = x^{ab}$ and $\frac {x^a}{x^b} = x^{a - b}$

simplify piece by piece, then recombine when done

3. Thanks Jhevon! I didn't see the ab cancle. So my final answer is $2b\sqrt{2ab}$

4. Originally Posted by Jonboy
Thanks Jhevon! I didn't see the ab cancle. So my final answer is $2b\sqrt{2ab}$
i don't like the fact that you have b's in different locations, i would write it as:

$2 \sqrt {2ab^3}$ or something like that

5. Is your way the most common? I just want that way.

6. Originally Posted by Jonboy
Is your way the most common? I just want that way.
i don't think it's that big a deal, it's just a preference. i think you've demonstrated the skills that the exercise was testing for