Dividing Square Roots?

[leahmorgan13]

So I've searched every place on the web for how to do this question (my own school lessons don't help much either) and cannot find anything. I've finally set out to search for some help myself. So here I am.

If you could provide an explanation on how to do this so I can find the correct answer, I will be forever thankful. I have answer choices and have worked it out with my parents billions of times and have not come up with one of my choices yet.

√ 9
-------
^6√9

It would read as: the square root of 9 over the square root of 9 to the 6th power, I think

Answer choices are as follows:
A: 9 and negative 2/3
B: 9 and negative 1/3
C: 9 and 1/2
D: 9 and 1/3

I tried to do 9/9 in one radicand and multiply 1/6? Not really sure how to get started or do this at all

Bonus:
If none of my explanation makes sense, the question and answers look like this: http://prntscr.com/5ta9py

 

[HallsofIvy]

So I've searched every place on the web for how to do this question (my own school lessons don't help much either) and cannot find anything. I've finally set out to search for some help myself. So here I am.

If you could provide an explanation on how to do this so I can find the correct answer, I will be forever thankful. I have answer choices and have worked it out with my parents billions of times and have not come up with one of my choices yet.

√ 9
-------
^6√9

It would read as: the square root of 9 over the square root of 9 to the 6th power, I think

No. That "6" in front of the roots sign is NOT a power. "The square root of 9 to the sixth power would be \(\displaystyle \sqrt{9^6}= 9^3= 729\).

What you have in the denominator would be read "the sixth root of 9". That is, the value of x such that \(\displaystyle x^6= 9\). Since 6= 2(3), \(\displaystyle \frac{\sqrt{9}}{\sqrt[6]{9}}= \frac{\sqrt[6]{9}\sqrt[6]{9}\sqrt[6]{9}}{\sqrt[6]{9}}= \sqrt[6]{9}\sqrt[6]{9}= \left(\sqrt[6]{9}\right)^2= \sqrt[3]{9}\).

This could also be done by writing the roots as fractional exponents. \(\displaystyle \sqrt{x}= x^{1/2}\), \(\displaystyle \sqrt[3]{x}= x^{1/3}\), and \(\displaystyle \sqrt[6]{x}= x^{1/6}\) so you can write the original problem as \(\displaystyle \frac{9^{1/2}}{9^{1/6}}= 9^{1/2- 1/6}= 9^{1/3}\).

Answer choices are as follows:
A: 9 and negative 2/3
B: 9 and negative 1/3
C: 9 and 1/2
D: 9 and 1/3

None of those answers, at written, is correct. "9 and 1/3" mean 9+ 1/3= 27/3+ 1/3= 28/3. The correct answer is \(\displaystyle 9^{1/3}\), "9 to the 1/3" power.

I tried to do 9/9 in one radicand and multiply 1/6? Not really sure how to get started or do this at all

You seem to misunderstand the difference between adding or multiplying two numbers and taking powers or roots of a number. If this is a course you are taking talk to your teacher about that.

Bonus:
If none of my explanation makes sense, the question and answers look like this:

http://prntscr.com/5ta9py

 

[Dale10101]

Or, if you prefer a brisk walk before breakfast

So I've searched every place on the web for how to do this question (my own school lessons don't help much either) and cannot find anything. I've finally set out to search for some help myself. So here I am.

If you could provide an explanation on how to do this so I can find the correct answer, I will be forever thankful. I have answer choices and have worked it out with my parents billions of times and have not come up with one of my choices yet.

√ 9
-------
^6√9

It would read as: the square root of 9 over the square root of 9 to the 6th power, I think

Answer choices are as follows:
A: 9 and negative 2/3
B: 9 and negative 1/3
C: 9 and 1/2
D: 9 and 1/3

I tried to do 9/9 in one radicand and multiply 1/6? Not really sure how to get started or do this at all

Bonus:
If none of my explanation makes sense, the question and answers look like this: http://prntscr.com/5ta9py

\[\begin{array}{l}
\frac{{{9^{\frac{1}{2}}}}}{{{9^{\frac{1}{6}}}}} = \frac{{\sqrt[2]{9}}}{{\sqrt[6]{9}}} = \frac{3}{{\sqrt[{(3)(2)}]{9}}} = \frac{3}{{\sqrt[{(3)}]{{\sqrt[{(2)}]{9}}}}} = \frac{3}{{\sqrt[{(3)}]{3}}}\\
= {\frac{3}{{{{\left( {\sqrt[{(3)}]{3}} \right)}^1}}}^{}}\frac{{{{\left( {\sqrt[{(3)}]{3}} \right)}^2}}}{{{{\left( {\sqrt[{(3)}]{3}} \right)}^2}}} = \frac{3}{1}\frac{{{{\left( {\sqrt[{(3)}]{3}} \right)}^2}}}{{{{\left( {\sqrt[{(3)}]{3}} \right)}^{(1 + 2)}}}} = \frac{3}{1}\frac{{{{\left( {\sqrt[{(3)}]{3}} \right)}^2}}}{{{{\left( {\sqrt[{(3)}]{3}} \right)}^3}}} = \frac{3}{1}\frac{{{{\left( {\sqrt[{(3)}]{3}} \right)}^2}}}{3} = {\left( {\sqrt[{(3)}]{3}} \right)^2} = \left( {\sqrt[{(3)}]{{{3^2}}}} \right) = \sqrt[3]{9} = {9^{\frac{1}{3}}}
\end{array}\]

 

[Ishuda]

What you want to search for is something like
powers exponents
Although not at the top of the list on some of the search engines, I think wikipedia is a nice read
http://en.wikipedia.org/wiki/Exponentiation

The basic things you need to know for this problem are
\(\displaystyle ^{a}\sqrt{x} = x^{\frac{1}{a}}\)
x^-a = 1/x^a
and
x^a x^b = x^a+b