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abbie1

  • 2 years ago

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  1. freewilly922
    • 2 years ago
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    Convert them to exponents and then divide. \[\sqrt[n]{a} = a^{1/n}\]

  2. freewilly922
    • 2 years ago
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    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
    • 2 years ago
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    Which would then be \[x^{\frac{1}{12}}=\sqrt[12]{x}\]

  4. freewilly922
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    Um. No. What is \[\frac{5}{12}-\frac{1}{6}\]?

  7. freewilly922
    • 2 years ago
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    You are forgetting to put them into common denoms. 5\12 - 1\6 = 5\12 -2\12 = 3\12 = 1\4

  8. freewilly922
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    Sure.

  11. gohangoku58
    • 2 years ago
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    index 12√x^5 x index 6√x doesn't it mean, \(\huge \sqrt[12]{x^5} \times\sqrt[6]x\)

  12. freewilly922
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    i hope to question was avtually to divide... its bit unclear....

  17. gohangoku58
    • 2 years ago
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    *actually

  18. freewilly922
    • 2 years ago
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    yes it was. I am working on the fact that it said to divide the two expressions.

  19. gohangoku58
    • 2 years ago
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    the 'x' in middle is throwing me off...

  20. gohangoku58
    • 2 years ago
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    can u confirm the question abbie ?

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