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iwanttogotostanford

  • one year ago

Simplify square root of negative 48.

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  1. anonymous
    • one year ago
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    Can you think of any perfect squares that are factors of 48?

  2. iwanttogotostanford
    • one year ago
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    no not from the top of my head

  3. iwanttogotostanford
    • one year ago
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    its \[\sqrt{-48}\]

  4. anonymous
    • one year ago
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    I understand, but there's a method to my madness. You're going to have to simplify this radical, so you'll need to do this. The first few prefect squares are 1, 4, 9, 16, 25, 36, 49, etc. What is the largest of these that is a factor of 48?

  5. iwanttogotostanford
    • one year ago
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    4? because 4x12 is 48

  6. anonymous
    • one year ago
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    That's right, but is there a larger one?

  7. iwanttogotostanford
    • one year ago
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    I'm really rusty on my times table facts

  8. anonymous
    • one year ago
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    Use a calculator if it helps

  9. iwanttogotostanford
    • one year ago
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    ok

  10. iwanttogotostanford
    • one year ago
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    24

  11. anonymous
    • one year ago
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    Sorry, 24 is not a perfect square. I listed them above. What's the largest one that is a factor of 48?

  12. iwanttogotostanford
    • one year ago
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    I'm confused a bit, wouldn't it be 4? because 9x9 is 81 and thats too big

  13. anonymous
    • one year ago
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    9 is a perfect square because 3 x 3 = 9 16 is a perfect square because 4 x 4 = 16 25 is a perfect square because 5 x 5 = 25 etc. Which of those listed numbers is the largest one that is a factor of 48? You don't need to square them.

  14. iwanttogotostanford
    • one year ago
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    36

  15. anonymous
    • one year ago
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    Excellent. 16 x 3 = 48.

  16. iwanttogotostanford
    • one year ago
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    now what?

  17. anonymous
    • one year ago
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    Now, we're going to use the rules of working with radicals to simplify. Your question is\[\sqrt{-48}\]Having identified the largest perfect square that is a factor of 48 we can rewrite as follows\[\sqrt{-48}=\sqrt{\left( 16 \right)\left( -1 \right)\left( 3 \right)}\]Understand what we did here?

  18. iwanttogotostanford
    • one year ago
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    yes

  19. anonymous
    • one year ago
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    Good. Now the rules of radicals say that we can write this as follows\[\sqrt{-48} = \sqrt{\left( 16 \right)\left( -1 \right)\left( 3 \right)} = \sqrt{16}\sqrt{-1}\sqrt{3}\]You OK with that?

  20. iwanttogotostanford
    • one year ago
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    ok

  21. anonymous
    • one year ago
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    Good. You know what the square root of 16 is? And the square root of -1?

  22. iwanttogotostanford
    • one year ago
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    yes its 4 but i don't know the square root of -1

  23. anonymous
    • one year ago
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    You haven't studied imaginary numbers?

  24. iwanttogotostanford
    • one year ago
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    no, I'm learning them right now thats why i need help

  25. anonymous
    • one year ago
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    Well imaginary numbers are based on the square root of -1. It is an imaginary number that is given the symbol i. In other words\[\sqrt{-1} = i\]

  26. anonymous
    • one year ago
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    So, you have \(\sqrt{16} \sqrt{-1} \sqrt{3}\). And you know the square root of 16 and the square root of -1. Just substitute them in.

  27. iwanttogotostanford
    • one year ago
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    these are my answer choices negative 4 square root of 3 4 square root of negative 3 4 i square root of 3 4 square root of 3 i

  28. anonymous
    • one year ago
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    \[\sqrt{16}\sqrt{-1}\sqrt{3} = ?\]

  29. iwanttogotostanford
    • one year ago
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    would it be b or d?

  30. anonymous
    • one year ago
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    \(\sqrt{16}=4\) and \(\sqrt{-1} = i\) and \(\sqrt{3}\) can't be simplified any further. What does that give you?

  31. iwanttogotostanford
    • one year ago
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    D!?

  32. anonymous
    • one year ago
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    Well, the individual parts of D are correct but they're in the wrong order. Should have the rational number first, then the imaginary number i, then the radical. What other choice meets this description?

  33. iwanttogotostanford
    • one year ago
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    C then?

  34. anonymous
    • one year ago
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    That's correct. It's the convention that we write the answer \(4i\sqrt{3}\) rather than \(4\sqrt{3}i\) or \(\sqrt{3}i4\) or any other combination.

  35. iwanttogotostanford
    • one year ago
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    ok, thank you i was confused but now i understand better

  36. anonymous
    • one year ago
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    You're welcome

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