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anonymous

  • one year ago

solve the equation using quadratic formula. 2x (sqaured) +13x=0

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  1. anonymous
    • one year ago
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    can some one please help ,e i dont get quadratic functions

  2. pooja195
    • one year ago
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    Find a GCF first do you see any?

  3. anonymous
    • one year ago
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    13?

  4. AbdullahM
    • one year ago
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    It says to solve by using the Quadratic Formula... What is the Quadratic Formula?

  5. anonymous
    • one year ago
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    "ax2 + bx + c = 0"

  6. pooja195
    • one year ago
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    Thats standered form

  7. AbdullahM
    • one year ago
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    |dw:1434816541031:dw|

  8. anonymous
    • one year ago
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    yeah but for this question thers no c

  9. pooja195
    • one year ago
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    so that would be 0 then :)

  10. anonymous
    • one year ago
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    but dont u have tp plug in the numbers? or is it when theres no c then the answer is equivalent to 0?

  11. pooja195
    • one year ago
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    Name your abc values

  12. anonymous
    • one year ago
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    a=2 b=13

  13. pooja195
    • one year ago
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    \[\huge~\rm~ax^2+bx+c=0\] \[\huge~\rm~2x^2+13x+0=0\]

  14. pooja195
    • one year ago
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    Now name the abc values

  15. anonymous
    • one year ago
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    a=2 b=13 c=0?

  16. AbdullahM
    • one year ago
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    Great job so far :)

  17. pooja195
    • one year ago
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    yes! :) \[\huge~\rm~~x=\frac{ -(13)~\pm \sqrt{(13)^2-4(2)(0)} }{ 2(2)}\]

  18. anonymous
    • one year ago
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    \[so.... x= -13 \pm \sqrt{169}\]

  19. anonymous
    • one year ago
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    and then i would subtract 8?

  20. anonymous
    • one year ago
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    i dont understnd now...

  21. pooja195
    • one year ago
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    \[\huge~\rm~~x=\frac{ -13~\pm \sqrt{169~x~1} }{ 4}\] \[\huge~\rm~~x=\frac{ -13~\pm \sqrt{169} }{ 4}\] Whats the square root of 169?

  22. anonymous
    • one year ago
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    13

  23. anonymous
    • one year ago
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    right?

  24. pooja195
    • one year ago
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    \[\huge~\rm~~x=\frac{ -13~\pm~13 \sqrt{1} }{ 4}\] now set up two equations and solve \[\huge~\rm~~x=\frac{ -13~+~13 \sqrt{1} }{ 4}\] \[\huge~\rm~~x=\frac{ -13~-~13 \sqrt{1} }{ 4}\]

  25. anonymous
    • one year ago
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    so it would be -13/2 and 0?

  26. anonymous
    • one year ago
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    those are my two answers correct?

  27. anonymous
    • one year ago
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    so would i do the same thing for 6x (sqaured) +2x=4?

  28. pooja195
    • one year ago
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    You would subtract the 4 \[\huge~\rm~6x^2+2x-4=0\] and go from that

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