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anonymous

  • one year ago

3x^2-x-4=0

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  1. misty1212
    • one year ago
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    they really got you factoring, don't they?

  2. anonymous
    • one year ago
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    yes i took algebra 3 and this is all we are doing is factoring

  3. misty1212
    • one year ago
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    algebra 3? i didnt know they had that many algebras?

  4. anonymous
    • one year ago
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    neither did i until i went to this school

  5. misty1212
    • one year ago
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    try \[(3x-4)(x+1)=0\]

  6. arindameducationusc
    • one year ago
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    You can use quadratic equation also.....

  7. misty1212
    • one year ago
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    sure but that is more complicated, my guess is since one factors, they all factor

  8. arindameducationusc
    • one year ago
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    I agree with Misty's factorisation...... that's more easy....

  9. misty1212
    • one year ago
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    do you know what to do once you have \[(3x-4)(x+1)=0\]?

  10. anonymous
    • one year ago
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    it worked thank you @misty1212

  11. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

  12. anonymous
    • one year ago
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    i have one more problem if you don't mind helping :)

  13. misty1212
    • one year ago
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    not at all happy to

  14. anonymous
    • one year ago
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    \[6x^2-13x+6=0\]

  15. misty1212
    • one year ago
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    bet we can factor this one too huh?

  16. anonymous
    • one year ago
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    yeah pretty sure . my problem is trying to figure out what plus what would equal them middle number , and that would also equal the last number if multiplied

  17. misty1212
    • one year ago
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    \[(2 x-3) (3 x-2) = 0\] seems to work

  18. arindameducationusc
    • one year ago
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  19. arindameducationusc
    • one year ago
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    use this b=-13 a=6 and c=6

  20. misty1212
    • one year ago
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    it is not that easy to figure out people claim they have methods, but the methods amount to trial an error anyways

  21. misty1212
    • one year ago
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    that is why i like to cheat makes life easier this is all donkey work anyways

  22. anonymous
    • one year ago
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    thank you so much for the help :)

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