abbie1 3 years ago .

1. freewilly922

Convert them to exponents and then divide. $\sqrt[n]{a} = a^{1/n}$

2. freewilly922

So for example $\sqrt[4]{x^3}\text{ divided by }\sqrt[3]{x^2}$ would be $\frac{x^{\frac{3}{4}}}{x^{\frac{2}{3}}}$ or $x^{\frac{3}{4}-\frac{2}{3}}$

3. freewilly922

Which would then be $x^{\frac{1}{12}}=\sqrt[12]{x}$

4. freewilly922

Sometimes the test wants you to give you answer in terms of a radical, then you might have to convert it back. If they will accept the exponent version, then that is the way to go.

5. freewilly922

yes, what is $\frac{x^{5/12}}{x^{1/6}}$ Just because they want it in radical form as the answer doesn't mean that you can't do all the work in exponent form and then reconvert. $\frac{x^a}{x^b}=x^{a-b}$ This seems easier to me than trying to figure out the radical forms.

6. freewilly922

Um. No. What is $\frac{5}{12}-\frac{1}{6}$?

7. freewilly922

You are forgetting to put them into common denoms. 5\12 - 1\6 = 5\12 -2\12 = 3\12 = 1\4

8. freewilly922

When you divide same base different exponents you subtract the exponents from one another. $\frac{aaaaa}{aa}=\frac{a^5}{a^2}$ $\frac{aaa}{1}\frac{aa}{aa}=aaa = a^3= a^{5-2}=a^3$

9. freewilly922

No, x^2/3 is wrong. Sorry. If you want to do it in radical form you can. $\frac{\sqrt[12]{x^5}}{\sqrt[6]{x}}$ can also be done realizing that $\sqrt[5]{a} = \sqrt[10]{a^2}$ so you can do it that way. You need to convert the index 6 to a twelve

10. freewilly922

Sure.

11. gohangoku58

index 12√x^5 x index 6√x doesn't it mean, $$\huge \sqrt[12]{x^5} \times\sqrt[6]x$$

12. freewilly922

It does say to divide the expressions though. $\frac{\sqrt[12]{x^5}}{\sqrt[6]{x}}\to\frac{\sqrt[12]{x^5}}{\sqrt[12]{x^2}}\to$ $\sqrt[12]{\frac{x^5}{x^2}}\to\sqrt[12]{x^{5-2}}\to\sqrt[12]{x^3}\to \sqrt[4]{\sqrt[3]{x^3}}\to\sqrt[4]{x}$

13. freewilly922

The main methods you need to know is that 1) $\sqrt[5]{x} = \sqrt[10]{x^2}$ 2) $\frac{\sqrt[12]{a}}{\sqrt[12]{b}}=\sqrt[12]{\frac{a}{b}}$ Yes the fourth root of x is the answer in radical form.

14. freewilly922

The reason why I suggested immediately to use your exponent conversion is that, if this is algII, you probably learned the exponent rules prior to this and can then use them rather than memorize another set of rules for radicals.

15. freewilly922

for questions like these, you need to be able to write down the 4 or so rules of exponents and the radical methods as well. Good luck and don't give up. exponents give a lot of people confusion.

16. gohangoku58

i hope to question was avtually to divide... its bit unclear....

17. gohangoku58

*actually

18. freewilly922

yes it was. I am working on the fact that it said to divide the two expressions.

19. gohangoku58

the 'x' in middle is throwing me off...

20. gohangoku58

can u confirm the question abbie ?