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anonymous

  • 5 years ago

What is the quadratic formula? What is it used for? Give an uncommon example.

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  1. amistre64
    • 5 years ago
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    this one?

  2. anonymous
    • 5 years ago
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    It's one of three my friend! lol.

  3. amistre64
    • 5 years ago
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    an uncommon example eh..... the quadratic formula is derived from a method know as "completing the square"; and is used to find the real roots of a quadratic expression (quadratic just being a fancy name for "squared").

  4. amistre64
    • 5 years ago
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    there are 2 named parts to the quadratic formula that I tend to get backwards; the determinate and the discriminate..they were haveing a sale on "d" words back then and so were stuck with them

  5. anonymous
    • 5 years ago
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    lol....ok

  6. amistre64
    • 5 years ago
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    the discriminte, if it so be called, is the part -b/2a and it gives you the axis of symmetry of the parabola.... to find the vertex, the hoghest or lowest point of the parabola; we use the vertex in the equation to come up with the "y" part...make sense so far?

  7. amistre64
    • 5 years ago
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    the other "d" word, the determinate perhaps, is a square root portion of the quad formula and it tells us whether there is: no roots, 1 root, or 2 roots.

  8. anonymous
    • 5 years ago
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    ummmm........clear as mud. But it's mainly me.

  9. anonymous
    • 5 years ago
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    So that would be the uncommon example?

  10. amistre64
    • 5 years ago
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    dunno about any "uncommon" examples really, still trying to figure out what that means... maybe its talking about a quadratic that aint easily factored.....

  11. amistre64
    • 5 years ago
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    if the square root portion has under its radical umbrella a negative #, there are no real roots; if it equals zero, then there is only 1 root, and if it is greater than zero, there are going to be 2 roots.

  12. amistre64
    • 5 years ago
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    the quad formula is: -b sqrt(b^2 - 4ac) ---- +- ------------- 2a 2a here you can see the two parts :) i shoulda prolly put that up first...

  13. anonymous
    • 5 years ago
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    No just an everyday example that might not be the first one you think of. Like, doesn't airplanes have something to do with the quadratic formula?

  14. anonymous
    • 5 years ago
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    SO that problem has two roots?

  15. amistre64
    • 5 years ago
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    that is not a problem ; that is the formula to determine the roots :) the standard quadratic equation is of the form: ax^2 +bx +c where a.b.and c are numbers...coefficients..

  16. anonymous
    • 5 years ago
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    Lol....wow.

  17. amistre64
    • 5 years ago
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    depending on the values of a,b,and c we can use the quadratic formula to determine its axis of symmetry, vertex coordinates, and all known solutions to y=0

  18. amistre64
    • 5 years ago
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    lets try to make up an equation and test the quad formula out.... and try to stay away from airplanes :)

  19. amistre64
    • 5 years ago
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    what are your favorite three numbers in the whole wide world... :)

  20. anonymous
    • 5 years ago
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    3?

  21. amistre64
    • 5 years ago
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    3 numbers.....yeah

  22. anonymous
    • 5 years ago
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    Oh wait.......it said provide a USEFUL example..... duh. Not uncommon, or whatever I wrote.

  23. anonymous
    • 5 years ago
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    7, 3, 35

  24. amistre64
    • 5 years ago
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    ok..7 3 and 35; lets put them into the hat....stir it up...and pick one...............3 ok, a = 3; try again................7; b = 7 .......and stir again for effect........we get a 35; c = 35

  25. amistre64
    • 5 years ago
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    3x^2 +7x -35...thats a quadratic formula...is it useful? dunno. but lets try out our formula.

  26. amistre64
    • 5 years ago
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    the first thing we should do is see if its worth trying to find any roots: sqrt((7)^2 - (4)(3)(-35)) if this is zero or above, were good to go. sqrt(49 + 420) = sqrt(469) its a keeper so that last part is gonna be: +- sqrt(469)/2(3) the axis of symmetry is gonna be: -b/2a = -7/2(3) = -7/6

  27. amistre64
    • 5 years ago
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    so we have a center for our parabola, its at x = -7/6 if we add and subtract the other part from this, we will know our solutions.

  28. anonymous
    • 5 years ago
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    Dear God my brain hurts.

  29. anonymous
    • 5 years ago
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    Any useful examples to share?

  30. amistre64
    • 5 years ago
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    (-7/6) + sqrt(469)/6 = about 2.4427

  31. amistre64
    • 5 years ago
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    this might be useful :) one example is the arch formed by throwing a baseball to your son.... you can determine its position with a quadratic equation.... but the "roots" of it are not that exciting :)

  32. anonymous
    • 5 years ago
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    An arch of throwing a baseball?

  33. amistre64
    • 5 years ago
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    the other root we got is about: -4.7761 you never seen a person throw a baseball around? it tends to go up, then comes back down into the golve of the oter person. makes for a nice arch while it flys thru the air

  34. anonymous
    • 5 years ago
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    Yeah! Gotcha! Now......what do you know about Pythogorean Therorem's......or whatever it is?

  35. amistre64
    • 5 years ago
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    i know that guy was a nutjob..... but his theorum is pretty helpful :)

  36. anonymous
    • 5 years ago
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    So was Pavlov. Ok......one sec...

  37. anonymous
    • 5 years ago
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    There are many applications for the Pythagorean Theorem.. can you give a specific example and show with calculations how the Pythagoras theorem applies?

  38. amistre64
    • 5 years ago
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    this is the pyth thrm....

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  39. amistre64
    • 5 years ago
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    the square of the sum of the legs of a right trianlge are equal to the square of the hypotenuse side.

  40. amistre64
    • 5 years ago
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    a^2 + b^2 = c^2

  41. anonymous
    • 5 years ago
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    OH.....and one more on quadratics too..

  42. anonymous
    • 5 years ago
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    Nice attachment, am I supposed to color it? JK!

  43. amistre64
    • 5 years ago
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    lol..... use it as a place mat ;)

  44. anonymous
    • 5 years ago
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    So that a^2 thing is the formula for that>?

  45. amistre64
    • 5 years ago
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    it is.... the names of the variables are unimportant, but that is a common rendition of it. another one is: x^2 + y^2 = r^2 and another one is: sin^2 + cos^2 = 1

  46. anonymous
    • 5 years ago
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    all for the sum of the legs of a right triangle?

  47. amistre64
    • 5 years ago
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    it can even be expanded to: c^2 = a^2 + b^2 -2ab cos(C)

  48. amistre64
    • 5 years ago
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    yep, since right triangle are about the easiest to use; its nice to have stuff you can use with them :)

  49. amistre64
    • 5 years ago
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    since the cos(90) = 0, that last part vanishes in the right trianlge and your left with the sum of the squares of the sides :)

  50. anonymous
    • 5 years ago
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    Ok......one more the quadratics thing.....

  51. amistre64
    • 5 years ago
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    ok....... I hope your writeing this stuff down :) I might forget it all by tomorrow :)

  52. anonymous
    • 5 years ago
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    Explain the four-steps for solving quadratic equations. Can any of these steps be eliminated? Can the order of these steps be changed? Would you add any steps to make it easier, or to make it easier to understand?

  53. anonymous
    • 5 years ago
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    No kidding.......writing away!

  54. amistre64
    • 5 years ago
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    ive seen this one before in here and it seems to be an odd question in general. the four steps? they might be talking about the 4 common ways that quads are facttored, but then it says to elinate one of them as tho they were all used together.....

  55. amistre64
    • 5 years ago
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    the steps I use are: 1) look at the problem 2) factor as needed 3) write down the answer 4) go to next problem?

  56. anonymous
    • 5 years ago
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    factor left side? does that make sense?

  57. amistre64
    • 5 years ago
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    doesnt clarify my quandry.....no.....

  58. amistre64
    • 5 years ago
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    id have to look up and google it to see what its talking about to give a good answer

  59. anonymous
    • 5 years ago
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    Hmmm..........ok.

  60. amistre64
    • 5 years ago
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    Maybe something like this: Step 1: Find two numbers that add together to get the middle term, and multiply to get the last term. Step 2: Rewrite the equation with those numbers: Step 3: Factor the first two and last two terms separately: Step 4: If you've done this correctly, your two new terms should have a clearly visible common factor.

  61. anonymous
    • 5 years ago
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    4) Go to the next problem? HA!

  62. amistre64
    • 5 years ago
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    :) yeah. there are many techniques used to factor, I have my favorites, but really, the steps you are looking for will be defined by the source you are using

  63. anonymous
    • 5 years ago
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    Can any be elimianted?

  64. anonymous
    • 5 years ago
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    OIC.

  65. anonymous
    • 5 years ago
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    *eliminated

  66. amistre64
    • 5 years ago
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    im gonna go with a no on that one....eliminating steps just sounds bad...and Ive built alot of stairs in my day :)

  67. anonymous
    • 5 years ago
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    Could the steps be rearranged?

  68. amistre64
    • 5 years ago
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    gonna have to go with a no again for that one, the steps tend to be in order of what you do to get to the next one.....still sounds like a bad idea to be rearranging them.

  69. anonymous
    • 5 years ago
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    Lol, ok....well you have been an AWESOME help up until this question anyway! Thanks so much. My brain is cramping up at this point. :)

  70. amistre64
    • 5 years ago
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    Ciao bella :)

  71. anonymous
    • 5 years ago
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    Lol, I may just write down "it sounds like a bad idea to me" just to see what kind of feedback i get!.

  72. amistre64
    • 5 years ago
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    ..works for me :)

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