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poopsiedoodle Group Title

For problems 1–5, simplify the expression. Answers written in decimal form will not be accepted. Each of these problems is worth 1 point. "v/" is a radical by the way ___ 1.v/96 ______ 2. 8 v/63x^5 ___________ 3. v/128x^5y^2 ___ 4. ^3v/32 ________ 5. ^3v/56x^14

  • 2 years ago
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  1. poopsiedoodle Group Title
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    So, I need them in radical form.

    • 2 years ago
  2. freewilly922 Group Title
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    Factor everything first and then apply your radical rules... for example \[\sqrt[3]{x^3} = x\]

    • 2 years ago
  3. poopsiedoodle Group Title
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    Problem is that I have no idea what you're saying. ._.

    • 2 years ago
  4. freewilly922 Group Title
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    Ok for example \[\sqrt[8]{256x^4}\] becomes, if you factor \[\sqrt[8]{2^{8}x^4}\] which then becomes \[2\sqrt[8]{x^4}=2 x^{\frac{4}{8}}=2x^{1/2}=2\sqrt{2}\]

    • 2 years ago
  5. freewilly922 Group Title
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    This uses the idea that \[\sqrt[a]{x} = x^{1/a}\]

    • 2 years ago
  6. poopsiedoodle Group Title
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    just wondering, would you mind using bigger font? I can't see the exponents. To do that, put what you are saying in the curly braces in \(\huge\text{}\.) but take out the . at the end.

    • 2 years ago
  7. poopsiedoodle Group Title
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    the ^ means exponents. like, 2^3 is 2 cubed.

    • 2 years ago
  8. poopsiedoodle Group Title
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    \(\Large\text{But again, I have no idea how to do this.}\)

    • 2 years ago
  9. freewilly922 Group Title
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    is ^3v/56x^14 supposed to be \[\bigg(\sqrt{56x^{14}}\bigg)^3\] then?

    • 2 years ago
  10. poopsiedoodle Group Title
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    So, \(\huge\sqrt[8]{256x^4}\) turns into \(\huge\sqrt[8]{2^{8}x^4}\) which turns into \(\huge2\sqrt[8]{x^4}=2 x^{\frac{4}{8}}=2x^{1/2}=2\sqrt{2}\) But how? \(\huge\text{:|}\)

    • 2 years ago
  11. freewilly922 Group Title
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    Sorry that should be \[2\sqrt{x}\]

    • 2 years ago
  12. poopsiedoodle Group Title
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    and no, it's supposed to be \[^{3}\sqrt{56x^{14}}\]

    • 2 years ago
  13. freewilly922 Group Title
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    ok the basic theorems that you need for these type of problems are that \[\sqrt[a]{x} = x^{\frac{1}{a}}\] So for example \[\large \sqrt[3]{x^{10}}\to x^{\frac{10}{3}}\to x^3x^{\frac{1}{3}}\to x^3\sqrt[3]{x}\]

    • 2 years ago
  14. poopsiedoodle Group Title
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    I understand the first half of your example equation, but not the second half.

    • 2 years ago
  15. freewilly922 Group Title
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    So any complicated radical you are given you can convert to exponents, use the rules of exponents shamelessly and then convert back to radical. So \[\large \sqrt[3]{56x^{14}}\to \sqrt[3]{(7)(8)x^{14}}\to 7^{1/3}8^{1/3}x^{14/3}\] This then becomes \[ 7^{1/3}8^{1/3}x^{14/3}\to 7^{1/3}(2^3)^{1/3}x^{12/3}x^{2/3}\] becomes \[7^{1/3}2x^{4}x^{2/3}\to 2x^4 7^{1/3}x^{1/3}\to 2x^4\sqrt[3]{7x}\]

    • 2 years ago
  16. poopsiedoodle Group Title
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    my brain is about to explode. e_o

    • 2 years ago
  17. freewilly922 Group Title
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    So you don't understand how \[x^{10/3} \to x^3x^{1/3}\]? This results from the fact that \[X^aX^b = X^{ab}\] and the reverse. So if you have \[x^{10/3}\] you have \[x^{9/3 + 1/3}\to x^{9/3}x^{1/3}\to x^3x^{1/3}\]

    • 2 years ago
  18. poopsiedoodle Group Title
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    again, might I suggest the larger font size? Maybe Large would be a good replacement for huge though.

    • 2 years ago
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