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ladiesman217

  • 2 years ago

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  1. ladiesman217
    • 2 years ago
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    \[3x^2=33x+24\]

  2. ladiesman217
    • 2 years ago
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    @ajprincess @nubeer @.Sam. can you help me please

  3. ajprincess
    • 2 years ago
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    \(3x^2=33x+24\) \(3x^2-33x+24=0\) Use quadratic formula. \[x=\frac{ -b\pm\sqrt{b^2-4ac} }{ 2a }\] the values of a, b and c can be found by comparing the equation \(3x^2-33x+24=0\) with the general equation \(ax^2+bx+c=0\). Does that help? @ladiesman217

  4. ladiesman217
    • 2 years ago
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    ya

  5. gohangoku58
    • 2 years ago
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    **-24 dividing the entire equation by 3 would simplify calculations

  6. ajprincess
    • 2 years ago
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    it is actually \(3x^2-33x-24=0\). am sorry abt the mistake

  7. ajprincess
    • 2 years ago
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    ya. first factor out 3 and then divide by 3 both sides. After that use quadratic formula:)

  8. ladiesman217
    • 2 years ago
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    the equation i got was\[\frac{ -(-3)\sqrt{3^{2}-4(3)(-24)} }{ 2*3 }\]

  9. ajprincess
    • 2 years ago
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    b=-33

  10. ladiesman217
    • 2 years ago
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    this is the correct equation right \[\frac{ -(-33)\sqrt{-33^{2}-4(3)(-24)} }{ 2*3 }\]

  11. ajprincess
    • 2 years ago
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    it will be \[\frac{ -(-33)\sqrt{(-33)^2-4*3*(-24)} }{ 2*3}\]

  12. ladiesman217
    • 2 years ago
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    i got\[\frac{ 11\pm \sqrt{153} }{ 2 }\]

  13. ajprincess
    • 2 years ago
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    ya right:)

  14. ajprincess
    • 2 years ago
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    good work:)

  15. ladiesman217
    • 2 years ago
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    thnx

  16. ajprincess
    • 2 years ago
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    welcome:)

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