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anonymous

  • 5 years ago

for this quadratic equation, does a=6,b=-3, and c=-2?

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  1. anonymous
    • 5 years ago
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    \[6x ^{2}-3=-2x\]

  2. anonymous
    • 5 years ago
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    no....b=2, c=-3

  3. anonymous
    • 5 years ago
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    Or should b or c be positive?

  4. anonymous
    • 5 years ago
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    once you put everything together on one side, you get: 6x^2+2x-3=0

  5. anonymous
    • 5 years ago
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    6x^2+2x-3, a=6 b=2 c=-3

  6. anonymous
    • 5 years ago
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    ohh because it has x with it?

  7. anonymous
    • 5 years ago
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    exactly :)

  8. anonymous
    • 5 years ago
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    think of the generic form of polynomials like that: ax^2+bx+c

  9. anonymous
    • 5 years ago
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    thanks! I'm going to try and solve it, would you mind checking it to see if I have the correct answer? :)

  10. anonymous
    • 5 years ago
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    yup

  11. anonymous
    • 5 years ago
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    \[4+ or - \sqrt{76}\over 12\] not sure if this can be simplified or not

  12. anonymous
    • 5 years ago
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    D = 4+72 = 76 > 0 --> 2 real number (-2 +- sqrt(76)) 4

  13. anonymous
    • 5 years ago
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    yup ur rite sorry im typo, denominator is 12

  14. anonymous
    • 5 years ago
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    that should be a -2, not a 4. remember the formula is [-b +/- sqrt(b^2-4ac)]/2a

  15. anonymous
    • 5 years ago
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    ohh ok, it is -2 instead of 4 though?

  16. anonymous
    • 5 years ago
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    what is b in this equation? its 2...so-b is -2

  17. anonymous
    • 5 years ago
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    in the original equation it is -2..but does it turn to positive 2 when we solve it?

  18. anonymous
    • 5 years ago
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    in the original is 2, coz its -b so it's -2

  19. anonymous
    • 5 years ago
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    ohhhh right. My mistake. thanks for clarifying!

  20. anonymous
    • 5 years ago
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    :)

  21. anonymous
    • 5 years ago
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    wait a sec..i tried to submit the answer and it says it's not in simplest form?

  22. anonymous
    • 5 years ago
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    76=4x19. You can pull out a 2 there and then factor a 2 from the whole numerator and then you end up with: [-1+/-sqrt19]/6

  23. anonymous
    • 5 years ago
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    Thanks!

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