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iloveboys

Solve: 0 = 6x^2 - 10x - 4.

  • one year ago
  • one year ago

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  1. tim538
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    you could factor that equation.

    • one year ago
  2. iloveboys
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    how

    • one year ago
  3. DoomDude
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    http://www.wolframalpha.com/input/?i=0+%3D+6x%5E2+-+10x+-+4

    • one year ago
  4. tim538
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    http://www.purplemath.com/modules/factquad.htm

    • one year ago
  5. iloveboys
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    x+1/3, x=2???

    • one year ago
  6. iloveboys
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    x i meant to put x=

    • one year ago
  7. DoomDude
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    rewrite that please

    • one year ago
  8. iloveboys
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    x=-1/3, x=2????

    • one year ago
  9. iloveboys
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    is that correct?

    • one year ago
  10. jim_thompson5910
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    You can factor, but let's use the quadratic equation to solve this \[\Large 0 = 6x^2 - 10x - 4\] \[\Large 6x^2 - 10x - 4 = 0\] \[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\] \[\Large x = \frac{-(-10)\pm\sqrt{(-10)^2-4(6)(-4)}}{2(6)}\] \[\Large x = \frac{10\pm\sqrt{100-(-96)}}{12}\] \[\Large x = \frac{10\pm\sqrt{196}}{12}\] \[\Large x = \frac{10+\sqrt{196}}{12} \ \text{or} \ x = \frac{10-\sqrt{196}}{12}\] \[\Large x = \frac{10+14}{12} \ \text{or} \ x = \frac{10-14}{12}\] \[\Large x = \frac{24}{12} \ \text{or} \ x = \frac{-4}{12}\] \[\Large x = 2 \ \text{or} \ x = -\frac{1}{3}\] So you are correct iloveboys, nice work

    • one year ago
  11. iloveboys
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    thank you

    • one year ago
  12. DoomDude
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    great answer man! provs

    • one year ago
  13. iloveboys
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    i have anoion... i have a graph liker this..

    • one year ago
  14. iloveboys
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    |dw:1337814199315:dw|the question is what quadratcic* equationis represtenesed below... than the graph

    • one year ago
  15. iloveboys
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    its somting l;ike that but more stright and more nice and even

    • one year ago
  16. iloveboys
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    answers: Non-factorable Trinomial Difference of Two Squares Not enough information Perfect Square Trinomial

    • one year ago
  17. jim_thompson5910
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    since the x intercepts are -2 and 2, we know that x = -2 and x = 2 are zeros So... x = -2 or x = 2 x + 2 = 0 or x - 2 = 0 (x + 2)(x - 2) = 0 x^2 - 4 = 0 Which is a difference of two squares since 4 is 2^2

    • one year ago
  18. iloveboys
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    okay thak u.. i was like whatttt.. lol

    • one year ago
  19. iloveboys
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    Part 1: Solve each of the quadratic equations below and describe what the solution(s) represent to the graph of each. Show your work to receive full credit. •y = x2 + 3x + 2. •y = x2 + 2x + 1. Part 2: Using complete sentences, answer the following questions about the two quadratic equations above. •Do the two quadratic equations have anything in common? If so, what?. •What makes y = x2 + 3x + 2 different from y = x2 + 2x + 1?.

    • one year ago
  20. jim_thompson5910
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    one moment

    • one year ago
  21. iloveboys
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    okay sory for all of this

    • one year ago
  22. jim_thompson5910
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    no it's fine, just going to type it up elsewhere and copy/paste

    • one year ago
  23. jim_thompson5910
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    Part 1 y = x^2 + 3x + 2 0 = x^2 + 3x + 2 x^2 + 3x + 2 = 0 (x+2)(x+1) = 0 x+2=0 or x+1=0 x=-2 or x=-1 So the solutions to y = x^2 + 3x + 2 are x=-2 or x=-1 ------------------------------------------------------- y = x^2 + 2x + 1 0 = x^2 + 2x + 1 x^2 + 2x + 1 = 0 (x+1)^2 = 0 x+1 = 0 x = -1 So the only solution to y = x^2 + 2x + 1 is x = -1

    • one year ago
  24. jim_thompson5910
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    let me know when you're ready for part 2

    • one year ago
  25. iloveboys
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    i am sorry i was try to understand it got it know go ahead when your ready

    • one year ago
  26. jim_thompson5910
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    alright, no worries, take all the time you need and ask about anything

    • one year ago
  27. iloveboys
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    thank you i got it know.

    • one year ago
  28. iloveboys
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    now

    • one year ago
  29. iloveboys
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    go ahead

    • one year ago
  30. jim_thompson5910
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    alright, just checking

    • one year ago
  31. jim_thompson5910
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    Part 2 Similarities: The two are similar in that they both are quadratic equations and graph parabolas. The two also open up in the same direction and have the same basic shape. The two quadratics are also factorable. ------------------------------------------------------------------ Differences: y = x^2 + 3x + 2 has 2 solutions or zeros y = x^2 + 2x + 1 has only 1 solution y = x^2 + 3x + 2 intersects the x-axis twice at 2 different spots y = x^2 + 2x + 1 only touches the x-axis at one spot only

    • one year ago
  32. iloveboys
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    Solve: x2 - 2x - 24 = 0. Answer x = -4, x = 6 x = 4, x = -6 x = -4, x = -6 x = 4, x = 6

    • one year ago
  33. jim_thompson5910
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    I already did this one, did you need to see the steps again?

    • one year ago
  34. iloveboys
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    im done for today and im thankful that you helped me so much... hen will you be on gain. and no sorry i didnt post it

    • one year ago
  35. jim_thompson5910
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    I should be on tomorrow

    • one year ago
  36. iloveboys
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    okay well thank you again!

    • one year ago
  37. jim_thompson5910
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    If I'm not on, you can email me or add me on msn or yahoo messenger either one works

    • one year ago
  38. jim_thompson5910
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    you're welcome

    • one year ago
  39. iloveboys
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    sounds good.. what is your email?

    • one year ago
  40. jim_thompson5910
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    jim_thompson5910@hotmail.com

    • one year ago
  41. iloveboys
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    ill porbley message you threw yahoo lol not msn lol

    • one year ago
  42. jim_thompson5910
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    my messenger screen names are just jim_thompson5910

    • one year ago
  43. jim_thompson5910
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    so same name

    • one year ago
  44. iloveboys
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    kkkk thank you byee can you repost theanser the question Solve: x2 - 2x - 24 = 0.

    • one year ago
  45. jim_thompson5910
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    oh sry, one sec

    • one year ago
  46. iloveboys
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    dont be sorry i told u not to.. and than it asked me agiana nd i cant find it lol

    • one year ago
  47. jim_thompson5910
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    lol alright, one sec while I type it up

    • one year ago
  48. iloveboys
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    ok

    • one year ago
  49. jim_thompson5910
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    Use the quadratic formula to solve for x \[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\] \[\Large x = \frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-24)}}{2(1)}\] \[\Large x = \frac{2\pm\sqrt{4-(-96)}}{2}\] \[\Large x = \frac{2\pm\sqrt{100}}{2}\] \[\Large x = \frac{2+\sqrt{100}}{2} \ \text{or} \ x = \frac{2-\sqrt{100}}{2}\] \[\Large x = \frac{2+10}{2} \ \text{or} \ x = \frac{2-10}{2}\] \[\Large x = \frac{12}{2} \ \text{or} \ x = \frac{-8}{2}\] \[\Large x = 6 \ \text{or} \ x = -4\] So the answer is choice A

    • one year ago
  50. iloveboys
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    thank you jim

    • one year ago
  51. jim_thompson5910
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    anytime

    • one year ago
  52. iloveboys
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    bye

    • one year ago
  53. jim_thompson5910
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    cya later

    • one year ago
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