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anonymous

  • one year ago

Can someone help me:(? Decide which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring. -b b2 - 4ac 2a Use the part of the quadratic formula that you chose above and find its value given the following quadratic equation: 4x2 + 6x + 2 = 0 (2 points)

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  1. anonymous
    • one year ago
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    this was another one i dint quite get

  2. pooja195
    • one year ago
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    The "discriminant" gives this information. It is the square root part of the quadratic formula. \[\sqrt{b^2-4ac}\]

  3. pooja195
    • one year ago
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    From here solve the equation

  4. pooja195
    • one year ago
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    Start by listing your abc values a=? b=? c=?

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

  6. anonymous
    • one year ago
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    4x^2+6x++2=0

  7. anonymous
    • one year ago
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    is that how it should be set up in order

  8. pooja195
    • one year ago
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    yes but we need to figure out the values using \[\huge~\rm~ax^2+bx+c=0\] That equation can you list just the NUMBERS not the signs of variables

  9. anonymous
    • one year ago
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    im confused

  10. arindameducationusc
    • one year ago
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    @pooja195 after this, can you solve my question? http://openstudy.com/study#/updates/55b787e4e4b0f0c1f13fd1a9

  11. pooja195
    • one year ago
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    Ok anything ^2 is the a value anything with an "x" is the b value anything with no varible is the c value now list the numbers :)

  12. anonymous
    • one year ago
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    so b^2 is A, -b is B, and C is 2a?

  13. pooja195
    • one year ago
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    wut? o-o wow you really are confused ok lets start from scratch \[\huge~\rm~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] This is the quadratic formula \[\huge~\rm~ax^2+bx+c=0\] This is standered form The discernment is The "discriminant" gives this information. It is the square root part of the quadratic formula. \[\sqrt{b^2-4ac}\] We need to solve \[\huge~\rm~4x^2 + 6x + 2 = 0\] Notice how it all matches up with the standered form equation? \[\huge~\rm~ax^2+bx+c=0\]

  14. anonymous
    • one year ago
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    i know, i said that already

  15. pooja195
    • one year ago
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    You have the numbers wrong a=4 b=6 c=2

  16. anonymous
    • one year ago
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    oh

  17. anonymous
    • one year ago
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    then why do we need just the numbers

  18. pooja195
    • one year ago
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    We need just the numbers so that we can solve using the discernment

  19. anonymous
    • one year ago
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    ok

  20. arindameducationusc
    • one year ago
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    @pooja195 I request you... okay after this

  21. pooja195
    • one year ago
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    \[\sqrt{b^2-4ac}\] So now we have our abc values \[\huge~\rm~ \sqrt{6^2-4(4)(2)}\] Solve that

  22. anonymous
    • one year ago
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    6^2-8 under a square root idk what

  23. anonymous
    • one year ago
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    i started with parenthesis and then multiplied 4 by (8)

  24. pooja195
    • one year ago
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    ok lets do this step by step whats 6^2?

  25. anonymous
    • one year ago
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    36

  26. pooja195
    • one year ago
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    ok leave that aside whats -4 times 4 ?

  27. anonymous
    • one year ago
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    -16

  28. pooja195
    • one year ago
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    Ok good -16 x 2=?

  29. anonymous
    • one year ago
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    -32

  30. pooja195
    • one year ago
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    good whats 36-32?

  31. anonymous
    • one year ago
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    4

  32. anonymous
    • one year ago
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    The 4ac

  33. anonymous
    • one year ago
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    4ac decides whether you get complex number or not.

  34. pooja195
    • one year ago
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    good now whats the \[\huge~\rm \sqrt4\]

  35. anonymous
    • one year ago
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    2

  36. pooja195
    • one year ago
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    @saseal you are confusing the asker we are already done with the problem.

  37. anonymous
    • one year ago
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    oh ok

  38. pooja195
    • one year ago
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    @kinserkara so whats your final answer?

  39. anonymous
    • one year ago
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    sry about that

  40. anonymous
    • one year ago
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    2a

  41. pooja195
    • one year ago
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    its ok :) @kinserkara now 2a it would just be x=2

  42. anonymous
    • one year ago
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    so the final answer is x=2

  43. pooja195
    • one year ago
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    yep

  44. anonymous
    • one year ago
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    ok thanks

  45. pooja195
    • one year ago
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    No problem :)

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