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poopsiedoodle

  • 2 years ago

Can someone help explain how to do quadratic equations with the quadratic formula? I'll give you a problem. See comments.

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  1. poopsiedoodle
    • 2 years ago
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    \[\huge 0 = -4(-4t^2 -8t -1)\]

  2. hba
    • 2 years ago
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    Well the first step is to compare your equation with the standard form and get a,b and c \[\huge\ ax^2+bx+c=0\]

  3. poopsiedoodle
    • 2 years ago
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    So, how would I do that with this one? Multiply the parentheses stuff by -4?

  4. hba
    • 2 years ago
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    Seeing your question i would suggest you to first divide your whole equation by -4 and then compare

  5. poopsiedoodle
    • 2 years ago
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    \[\Huge 0=-4t^2 -8t-1\]

  6. hba
    • 2 years ago
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    What is a,b and c in this ?

  7. poopsiedoodle
    • 2 years ago
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    a=-4 b=-8 c=-1

  8. hba
    • 2 years ago
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    Good,Now you can use the quad formula. \[\huge\ x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

  9. hba
    • 2 years ago
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    The other think which you can do with your equation is that you can take -1 common and then compare so everything becomes positive.you can also do it straight away :)

  10. poopsiedoodle
    • 2 years ago
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    so, t = -8*√64-4*-4 ------------------ 2*-4 ?

  11. poopsiedoodle
    • 2 years ago
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    which would then be t=-8*(8+4) ----------- -8

  12. poopsiedoodle
    • 2 years ago
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    am I getting it right?

  13. hba
    • 2 years ago
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    It should be +- not *

  14. poopsiedoodle
    • 2 years ago
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    I don't get it... wouldn't you be multiplying -8 and \[\pm √64-4*-4\]?

  15. hba
    • 2 years ago
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    t=-8(+-)(8+4) ----------- -8

  16. poopsiedoodle
    • 2 years ago
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    Now I'm rather confused.

  17. hba
    • 2 years ago
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    Lemme show you.

  18. hba
    • 2 years ago
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    I appreciate you working with me though.

  19. poopsiedoodle
    • 2 years ago
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    I guess I could do that \(\Huge\mathsf{:P}\)

  20. hba
    • 2 years ago
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    a=-4 b=-8 c=-1 \[\huge\ t=\frac{ -(-8) \pm \sqrt{(-8)^2-4(-4)(-1)} }{ 2(-4) }\]

  21. poopsiedoodle
    • 2 years ago
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    Ok, but what does the \[\pm\] do? what's its significance?

  22. hba
    • 2 years ago
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    Well when we simplify it we would be taking the adding part on one side and subracting part on the other side.

  23. poopsiedoodle
    • 2 years ago
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    so it'd be 8+√blahblahblah --------------- 2-(-4)?

  24. hba
    • 2 years ago
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    Either, 8+√blahblahblah --------------- 2-(-4)? Or, 8-√blahblahblah --------------- 2-(-4)?

  25. poopsiedoodle
    • 2 years ago
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    so the \[\pm\] just breaks up between the denominator and the numerator?

  26. hba
    • 2 years ago
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    Exactly.

  27. poopsiedoodle
    • 2 years ago
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    Ah. I guess I'll try another problem here and ask you if it's right or something.

  28. hba
    • 2 years ago
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    Sure :)

  29. poopsiedoodle
    • 2 years ago
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    \[\huge 0=-2x ^{2} +14x +20\] \[\large x=\frac{ -14\pm \sqrt{14^2 - 4(-2+20)} }{ -4 }\] \[\large x= \frac{ -14+14-72 }{ -4 }\]

  30. poopsiedoodle
    • 2 years ago
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    and the answer is \[\huge \frac{ -7\pm \sqrt{89} }{ 2 }\]. Would I be right?

  31. hba
    • 2 years ago
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    Its is -4(a)(c) not -4(a+c)

  32. poopsiedoodle
    • 2 years ago
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    oh. Well that ruins that then.

  33. hba
    • 2 years ago
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    Yes it does :) Do it on a paper and tell me what you got from that equation :)

  34. poopsiedoodle
    • 2 years ago
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    \[\Large x=\frac{ -14\pm \sqrt{14^2 - 4(-2*20)} }{ -4 }\] \[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\] \[\Large x=\frac{ -14\pm 14-160} { -4}\] \[\Large x=\frac{ -14\pm -146} {-4}\] set x to 0 and divide all by -2 to get \[\Large \frac{ 7\pm 73} { 2}\]

  35. poopsiedoodle
    • 2 years ago
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    and now I'm confused.

  36. poopsiedoodle
    • 2 years ago
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    oh wait, I forgot to keep the x's in there. Crap.

  37. poopsiedoodle
    • 2 years ago
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    Think you could help me out here? My mind is about to implode.

  38. hba
    • 2 years ago
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    x=7+73/2,x=7-73/2

  39. hba
    • 2 years ago
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    Btw, (-) *(-)=+

  40. hba
    • 2 years ago
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    So just because of signs it goes wrong.

  41. hba
    • 2 years ago
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    \[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\] Continue from here.

  42. poopsiedoodle
    • 2 years ago
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    But I left the xs out of my work. ._.

  43. hba
    • 2 years ago
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    No worries about that :)

  44. hba
    • 2 years ago
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    Continue working from here.

  45. poopsiedoodle
    • 2 years ago
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    \[\Large x=\frac{ -14\pm 14 - 4(-40)} { -4 }\] \[\Large x=\frac{ -14\pm 14+160} {-4}\] \[\Large x=\frac{ -14\pm 174} {-4}\] \[\Large x=\frac {\frac{ -14\pm 174} {-4}} {-2}\] \[\Large x=\frac {7\pm -87} {2}\]

  46. poopsiedoodle
    • 2 years ago
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    And now what do I do?

  47. hba
    • 2 years ago
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    I don't know what you are upto ,where did that divide by -2 come in between ?

  48. poopsiedoodle
    • 2 years ago
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    Well, I'm not sure. I just thought I had to do that. I'm rather confused ._.

  49. hba
    • 2 years ago
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    Hmm,If OS would be working on chrome i would have worked one problem for you. Can we do this later ?

  50. poopsiedoodle
    • 2 years ago
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    I'm using chrome right now o_o But, I guess that'd be OK :P

  51. hba
    • 2 years ago
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    Thanks.

  52. poopsiedoodle
    • 2 years ago
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    Thanks to you too :3

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