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AmTran_Bus

  • one year ago

quad formula help

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  1. AmTran_Bus
    • one year ago
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    |dw:1440200122850:dw|

  2. AmTran_Bus
    • one year ago
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    does the two also apply to what is under the root?

  3. AmTran_Bus
    • one year ago
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    So is it|dw:1440200448959:dw|

  4. arindameducationusc
    • one year ago
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    wait you made an error

  5. AmTran_Bus
    • one year ago
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    help me man by book says it is z=1+ or - sqrt (-4) or z=1 + or - 2i

  6. arindameducationusc
    • one year ago
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    |dw:1440200560042:dw|

  7. arindameducationusc
    • one year ago
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    do you know the defination of 'i'?

  8. arindameducationusc
    • one year ago
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    i=sqrt(-1)

  9. AmTran_Bus
    • one year ago
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    Yea.

  10. AmTran_Bus
    • one year ago
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    Can you show more steps

  11. AmTran_Bus
    • one year ago
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    Like I know the 2 cancels but what about the rest

  12. AmTran_Bus
    • one year ago
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    @arindameducationusc

  13. arindameducationusc
    • one year ago
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    yes yes have patience..

  14. jdoe0001
    • one year ago
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    \(\bf \textit{quadratic formula}\\ {\color{blue}{ 1}}z^2{\color{red}{ -2}}z{\color{green}{ +5}}=0 \qquad \qquad z= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\) what does that give you?

  15. radar
    • one year ago
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    The denominator "2" applies to both terms of the numerator.

  16. AmTran_Bus
    • one year ago
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    @radar exactly what I needed. So it is actually 1+ sqrt(-4) which is 1+ 2i

  17. arindameducationusc
    • one year ago
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    |dw:1440200937959:dw| simplify these to get your answer. hope this helps

  18. radar
    • one year ago
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    Yes. Note there are two complex solutions

  19. AmTran_Bus
    • one year ago
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    1+ or - 2i

  20. radar
    • one year ago
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    Correcto

  21. AmTran_Bus
    • one year ago
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    thanks a million

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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