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iloveboys Group Title

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

  • 2 years ago
  • 2 years ago

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  1. tim538 Group Title
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    you could factor that equation.

    • 2 years ago
  2. iloveboys Group Title
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    how

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

    • 2 years ago
  4. tim538 Group Title
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    http://www.purplemath.com/modules/factquad.htm

    • 2 years ago
  5. iloveboys Group Title
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    x+1/3, x=2???

    • 2 years ago
  6. iloveboys Group Title
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    x i meant to put x=

    • 2 years ago
  7. DoomDude Group Title
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    rewrite that please

    • 2 years ago
  8. iloveboys Group Title
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    x=-1/3, x=2????

    • 2 years ago
  9. iloveboys Group Title
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    is that correct?

    • 2 years ago
  10. jim_thompson5910 Group Title
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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

    • 2 years ago
  11. iloveboys Group Title
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    thank you

    • 2 years ago
  12. DoomDude Group Title
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    great answer man! provs

    • 2 years ago
  13. iloveboys Group Title
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    i have anoion... i have a graph liker this..

    • 2 years ago
  14. iloveboys Group Title
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    |dw:1337814199315:dw|the question is what quadratcic* equationis represtenesed below... than the graph

    • 2 years ago
  15. iloveboys Group Title
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    its somting l;ike that but more stright and more nice and even

    • 2 years ago
  16. iloveboys Group Title
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    answers: Non-factorable Trinomial Difference of Two Squares Not enough information Perfect Square Trinomial

    • 2 years ago
  17. jim_thompson5910 Group Title
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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

    • 2 years ago
  18. iloveboys Group Title
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    okay thak u.. i was like whatttt.. lol

    • 2 years ago
  19. iloveboys Group Title
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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?.

    • 2 years ago
  20. jim_thompson5910 Group Title
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    one moment

    • 2 years ago
  21. iloveboys Group Title
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    okay sory for all of this

    • 2 years ago
  22. jim_thompson5910 Group Title
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    no it's fine, just going to type it up elsewhere and copy/paste

    • 2 years ago
  23. jim_thompson5910 Group Title
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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

    • 2 years ago
  24. jim_thompson5910 Group Title
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    let me know when you're ready for part 2

    • 2 years ago
  25. iloveboys Group Title
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    i am sorry i was try to understand it got it know go ahead when your ready

    • 2 years ago
  26. jim_thompson5910 Group Title
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    alright, no worries, take all the time you need and ask about anything

    • 2 years ago
  27. iloveboys Group Title
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    thank you i got it know.

    • 2 years ago
  28. iloveboys Group Title
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    now

    • 2 years ago
  29. iloveboys Group Title
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    go ahead

    • 2 years ago
  30. jim_thompson5910 Group Title
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    alright, just checking

    • 2 years ago
  31. jim_thompson5910 Group Title
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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

    • 2 years ago
  32. iloveboys Group Title
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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

    • 2 years ago
  33. jim_thompson5910 Group Title
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    I already did this one, did you need to see the steps again?

    • 2 years ago
  34. iloveboys Group Title
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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

    • 2 years ago
  35. jim_thompson5910 Group Title
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    I should be on tomorrow

    • 2 years ago
  36. iloveboys Group Title
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    okay well thank you again!

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

    • 2 years ago
  38. jim_thompson5910 Group Title
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    you're welcome

    • 2 years ago
  39. iloveboys Group Title
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    sounds good.. what is your email?

    • 2 years ago
  40. jim_thompson5910 Group Title
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    jim_thompson5910@hotmail.com

    • 2 years ago
  41. iloveboys Group Title
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    ill porbley message you threw yahoo lol not msn lol

    • 2 years ago
  42. jim_thompson5910 Group Title
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    my messenger screen names are just jim_thompson5910

    • 2 years ago
  43. jim_thompson5910 Group Title
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    so same name

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

    • 2 years ago
  45. jim_thompson5910 Group Title
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    oh sry, one sec

    • 2 years ago
  46. iloveboys Group Title
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    dont be sorry i told u not to.. and than it asked me agiana nd i cant find it lol

    • 2 years ago
  47. jim_thompson5910 Group Title
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    lol alright, one sec while I type it up

    • 2 years ago
  48. iloveboys Group Title
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    ok

    • 2 years ago
  49. jim_thompson5910 Group Title
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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

    • 2 years ago
  50. iloveboys Group Title
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    thank you jim

    • 2 years ago
  51. jim_thompson5910 Group Title
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    anytime

    • 2 years ago
  52. iloveboys Group Title
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    bye

    • 2 years ago
  53. jim_thompson5910 Group Title
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    cya later

    • 2 years ago
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