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anonymous

  • 5 years ago

Could anyone please help me with simplifying radicals? -15

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  1. anonymous
    • 5 years ago
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    √(-15) is the problem?

  2. anonymous
    • 5 years ago
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    \[\sqrt{-15}\]

  3. anonymous
    • 5 years ago
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    what is your question?

  4. anonymous
    • 5 years ago
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    it is odd and cannot be simplfied

  5. anonymous
    • 5 years ago
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    ok so if you break it into it's factors you get √(-1)(3)(5) √(-1) = i the rest cannot be simplified so you have i√15 for an answer

  6. anonymous
    • 5 years ago
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    √49 is also odd but simplifies to 7 mmolina

  7. anonymous
    • 5 years ago
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    so the answer is -1/15

  8. anonymous
    • 5 years ago
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    ?

  9. anonymous
    • 5 years ago
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    no, the answer is: i*srqr(15)

  10. anonymous
    • 5 years ago
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    i squared as in india type of i?

  11. anonymous
    • 5 years ago
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    \[i*\sqrt{15}\]

  12. anonymous
    • 5 years ago
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    hmm never heard of that before. what rule is that derived from?

  13. anonymous
    • 5 years ago
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    i is the imaginary unit √(-1) because you cannot take the square root of a negative number.

  14. anonymous
    • 5 years ago
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    okay. let me try that...

  15. anonymous
    • 5 years ago
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    set in you personal calculator \[\sqrt{-1}\] this will give you an error as the answer. So every time you have an expresion like that you have to write an i in front of the root and make the number inside the root positive.

  16. anonymous
    • 5 years ago
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    the entire problem is quadratic formulas which is -1 +/- \[\sqrt{-15}/2\]

  17. anonymous
    • 5 years ago
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    then there are no real solutions

  18. anonymous
    • 5 years ago
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    I have the left side down and the fraction beneathe, but complicated fixing right side. is that possible for a solution set? the answer must be two separate numbers

  19. anonymous
    • 5 years ago
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    not always, quadratics have AT MOST 2 solutions but can have only one or no real solutions, this appears to be a classic no real solutions problem

  20. anonymous
    • 5 years ago
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    the solutions to a quadratic are where the graph crossed the x-axis, if you graph the original problem does it cross the x-axis?

  21. anonymous
    • 5 years ago
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    there is no graphing, and there is no option for no solution--there is indeed a solution. haha

  22. anonymous
    • 5 years ago
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    It might make things simpler if you tell us what the original problem is.

  23. anonymous
    • 5 years ago
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    it must form an equation with only addition and subraction signs between them

  24. anonymous
    • 5 years ago
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    I did if you look above

  25. anonymous
    • 5 years ago
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    :)

  26. anonymous
    • 5 years ago
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    hello?

  27. anonymous
    • 5 years ago
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    did you not understand it?

  28. anonymous
    • 5 years ago
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    I understand that you have simplified the original problem but if you post what the quadratic is then we could verify that your solutions are correct.

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