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karinewoods17

  • 2 years ago

combine {sqrt -8} + {sqrt -32}

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  1. karinewoods17
    • 2 years ago
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    The answer in my book is \[6i - 2i \sqrt{10}\] I dont know how they got that answer.

  2. ilikephysics2
    • 2 years ago
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    You can't take the square root of a negative number

  3. ChmE
    • 2 years ago
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    @ksaimouli is incorrect

  4. karinewoods17
    • 2 years ago
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    thats why you multiply it by i which is equal to -1

  5. Ahaanomegas
    • 2 years ago
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    @ksaimouli That's totally wrong. sqrt(a)+sqrt(b) is NOT sqrt(a+b). @ilikephysics2 Yes, You can. That's what imaginary numbers are there for. @karinewoods17 Notice that \[ \Large {\sqrt {8} = 2\sqrt{2}} \]and \[ \Large {\sqrt {32} = 4\sqrt{2}}, \]so \[ \Large {\sqrt {-8} = 2i\sqrt{2}} \]and \[ \Large {\sqrt {-32} = 4i\sqrt{2}} \]Do you see how that works? Now, just add to get 6isqrt(2).

  6. karinewoods17
    • 2 years ago
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    wouldnt it be \[6i \sqrt{2}\] ??

  7. ChmE
    • 2 years ago
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    \[\sqrt{-8}+\sqrt{-32}=\sqrt{-8}+\sqrt{-8\times4}= \sqrt{-8} + 2\sqrt{-8}\]

  8. ChmE
    • 2 years ago
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    \[3\sqrt{-8}=3i \sqrt{8}=3i \sqrt{2\times4}=6i \sqrt{2}\]

  9. ChmE
    • 2 years ago
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    Yes you are correct

  10. Ahaanomegas
    • 2 years ago
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    Yep. That's exactly what I said, except I explained everything. =P

  11. karinewoods17
    • 2 years ago
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    but... the answer in my book says the answer is \[6i - 2i \sqrt{10}\] How did they get that?

  12. karinewoods17
    • 2 years ago
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    Im sorry:/ Im making coffee and waiting for you:)

  13. karinewoods17
    • 2 years ago
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    im not! sorry!

  14. karinewoods17
    • 2 years ago
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    I was making breakfast for my boyfriend.

  15. karinewoods17
    • 2 years ago
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    umm. \[\sqrt{-32}\] breaks into \[\sqrt{-32} = \sqrt{4} * \sqrt{-8}\] which simplifies to \[4i \sqrt{2}\] ?

  16. karinewoods17
    • 2 years ago
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    And yes I try:)

  17. karinewoods17
    • 2 years ago
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    now what. lol

  18. karinewoods17
    • 2 years ago
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    @nincompoop please do not ignore me;)

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