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gahm8684

  • one year ago

Imaginary complex numbers

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  1. zzr0ck3r
    • one year ago
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    THEY ARE EVERYWHERE!!!!!

  2. gahm8684
    • one year ago
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    I've been asked to solve this equation, however when I get to certain step I get stuck and don't know what to do \[x^2+2x+3=0\]

  3. zzr0ck3r
    • one year ago
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    What step?

  4. gahm8684
    • one year ago
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    \[(-2\pm \sqrt{4-12})/2\]

  5. gahm8684
    • one year ago
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    my bad I meant \[-2\pm \sqrt{-8}/2\]

  6. zzr0ck3r
    • one year ago
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    \[\sqrt{4-12}=\sqrt{-8}=\sqrt{-1*8}=\sqrt{-1}\sqrt{8}=i\sqrt{8}=i\sqrt{4*2}=i\sqrt{4}\sqrt{2}=2i\sqrt{2}\]

  7. gahm8684
    • one year ago
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    ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooohohohhohhohoohohohohohohohoohohohohohhhhhhhhhhhhhhhh

  8. zzr0ck3r
    • one year ago
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    Yoou should end up with \(-1\pm i\sqrt{2}\)

  9. gahm8684
    • one year ago
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    I get it.

  10. zzr0ck3r
    • one year ago
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    word

  11. gahm8684
    • one year ago
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    Thank you

  12. zzr0ck3r
    • one year ago
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    np

  13. gahm8684
    • one year ago
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    How did you get the -1 tho

  14. zzr0ck3r
    • one year ago
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    \(-8=-1*8\)

  15. gahm8684
    • one year ago
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    no, I'm talking about the one in -1pmisqrt(2)

  16. zzr0ck3r
    • one year ago
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    \(\dfrac{-2\pm2i\sqrt{2}}{2}=\dfrac{-2}{2}\pm \dfrac{2i\sqrt{2}}{2}=-1\pm i\sqrt{2}\)

  17. gahm8684
    • one year ago
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    what about the 2i, what happened to the 2?

  18. zzr0ck3r
    • one year ago
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    \[\dfrac{-2}{2}\pm\dfrac{2i\sqrt{2}}{2}=\dfrac{\cancel{-2}}{\cancel{2}}\pm\dfrac{\cancel{2}i\sqrt{2}}{\cancel{2}}=-1\pm i\sqrt{2}\]

  19. zzr0ck3r
    • one year ago
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    \[\dfrac{ab+cb}{b}=a+c\]

  20. gahm8684
    • one year ago
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    alright I get it now thank you

  21. zzr0ck3r
    • one year ago
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    np

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