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anonymous

  • 5 years ago

3x^2-7x-3=0

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  1. amistre64
    • 5 years ago
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    3times3 is 9 which doesnt factor into 7.... so quadratic it :)

  2. anonymous
    • 5 years ago
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    after using the quadratic formula, how do you complete the problem?

  3. amistre64
    • 5 years ago
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    that is the completed problem. you could try doing the "complete the square" method; but all that method does is prove that the quad formula works.

  4. anonymous
    • 5 years ago
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    my textbook says to find the nearest degree.. x = sin of theta

  5. amistre64
    • 5 years ago
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    then find what x equals and take the sine inverse of it. the inverse trig funtions take numbers for inputs and spit back angles. Does that make sense?

  6. anonymous
    • 5 years ago
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    not really :/

  7. amistre64
    • 5 years ago
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    lets walk you thru it then :) first get the values for x with the quad formula. 7/6 (+-) sqrt(49-(4)(-3)(3))/ 6 7/6 (+-) sqrt(49+36)/6 7/6 (+-) sqrt(85)/6 you with me so far?

  8. anonymous
    • 5 years ago
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    ok, let me see.

  9. anonymous
    • 5 years ago
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    where do you get 49?

  10. amistre64
    • 5 years ago
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    the quad form reads: -middle (+-) sqrt(middle^2 -4(first)(last)) all over 2(first). or more formally: -b (+-) sqrt(b^2-4(a)(c)) --------------------- 2(a) right? -7 times -7 = 49

  11. anonymous
    • 5 years ago
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    oh. it's b^2? no wonder i never got any solution out of that. thank you very much! +5

  12. amistre64
    • 5 years ago
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    we got numbers for our x now; (7/6 (+-) sqrt(85)/6) if I did it correctly right?

  13. anonymous
    • 5 years ago
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    umm.. i'm still solving it. i got the b^2 correct

  14. anonymous
    • 5 years ago
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    my calculator says -49 instead of 49. so weird~

  15. amistre64
    • 5 years ago
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    lol.... maybe user error? To find out what sin(t) is when x equals some number we write it like this: sin(t) = x To find out what "t" is we need to get rid of the (sin) part. And we do that by using what is called the sine inverse function.

  16. anonymous
    • 5 years ago
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    i think it was supposed to be 7 not -7

  17. amistre64
    • 5 years ago
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    all it changes in the end is the sign of (7/6) to (-7/6) which may or may not be an issue

  18. anonymous
    • 5 years ago
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    got it. \[x=7\pm \sqrt{49+36} \div 6\]

  19. amistre64
    • 5 years ago
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    if your equation you gave was correct, then it was supposed to be (-7)^2 = 49. But (7)^2 = 49 just the same

  20. amistre64
    • 5 years ago
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    good :) you ready to learn about sin inverse? its symbol is (sin^-1)

  21. anonymous
    • 5 years ago
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    i know sin^-1 but i'm stuck in where should i go next.

  22. amistre64
    • 5 years ago
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    lets walk it thru. It asks us to find sin(t) = x right?

  23. anonymous
    • 5 years ago
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    sin \[\theta\]

  24. amistre64
    • 5 years ago
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    theta is simply a name, I will make it simpler and just name it "t" do you think I can do that?

  25. anonymous
    • 5 years ago
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    yea, sure. :)

  26. amistre64
    • 5 years ago
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    lol...good, mathmaticians like to rename things all the time, makes them feel important :)

  27. anonymous
    • 5 years ago
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    XD

  28. amistre64
    • 5 years ago
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    we have: sin(t) = x we need to get rid of the "sin" part. So lets apply sin^-1 to each side to get: t = sin^-1(x) Now plug the numbers into your scientific calculator to get an angle for t, formally known as theta

  29. anonymous
    • 5 years ago
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    wait. i'm confused here. do i continue with the quad. formula?

  30. amistre64
    • 5 years ago
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    the quad formula simply gave us the values for x....its job is done and it can go take a well deserved break...

  31. anonymous
    • 5 years ago
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    so i continue with this:\[3\sin^2\theta-7\sin \theta-3=0\]

  32. amistre64
    • 5 years ago
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    aha!!! you were holding back vital information...hmmm

  33. anonymous
    • 5 years ago
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    opps.. am i in trouble?

  34. amistre64
    • 5 years ago
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    nah....just makes the problem more understandable now :)

  35. anonymous
    • 5 years ago
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    XD. so how do i continue?

  36. amistre64
    • 5 years ago
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    just the same as before :) sin(t) = x and solve for t But I gotta tell wonder, do you have the problem written correctly becuase I get some wierd stuff when I plug it into the calculator..

  37. anonymous
    • 5 years ago
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    the textbook says to use the quadratic formula to find,to the nearest degree, all values of theta in the interval 0 degrees [less than or equal to] theta [less than] 360 degrees that satisfy each equation.

  38. amistre64
    • 5 years ago
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    338.29 degrees is what I get, which is in Quad4 but it is also hiding in Quad3 under the disguise of 201.71 degrees.

  39. amistre64
    • 5 years ago
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    [7+sqrt(85)] /6 is greater then 1 so it spits out an error. x = 2.7 and that aint between -1 and 1 so it is a false reading.

  40. amistre64
    • 5 years ago
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    [7-sqrt(85)]/6 = -0.3699 which IS between -1 and 1 so it is a good value of x.

  41. amistre64
    • 5 years ago
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    is there anyway to verify the answer?

  42. anonymous
    • 5 years ago
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    i got the same answer as you. -0.3699240762

  43. amistre64
    • 5 years ago
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    either were both geniuses or were both idiots...take your pick :)

  44. anonymous
    • 5 years ago
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    GENIUS! :D

  45. amistre64
    • 5 years ago
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    yay!!!

  46. amistre64
    • 5 years ago
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    the thing about sin^-1 is that it only gives an angle that is on the right hand side of a circle. So you have to remember to look over to the other sid eto see where it matches up at.

  47. amistre64
    • 5 years ago
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    does any of that make sense to you?

  48. anonymous
    • 5 years ago
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    somewhat, let me think over~

  49. anonymous
    • 5 years ago
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    i think i got it, thanks for helping me though(:

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