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iloveboys

  • 2 years ago

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

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  1. tim538
    • 2 years ago
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    you could factor that equation.

  2. iloveboys
    • 2 years ago
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    how

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

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

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

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

  7. DoomDude
    • 2 years ago
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    rewrite that please

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

  9. iloveboys
    • 2 years ago
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    is that correct?

  10. jim_thompson5910
    • 2 years ago
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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

  11. iloveboys
    • 2 years ago
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    thank you

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

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

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

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

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

  17. jim_thompson5910
    • 2 years ago
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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

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

  19. iloveboys
    • 2 years ago
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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?.

  20. jim_thompson5910
    • 2 years ago
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    one moment

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

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

  23. jim_thompson5910
    • 2 years ago
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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

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

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

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

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

  28. iloveboys
    • 2 years ago
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    now

  29. iloveboys
    • 2 years ago
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    go ahead

  30. jim_thompson5910
    • 2 years ago
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    alright, just checking

  31. jim_thompson5910
    • 2 years ago
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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

  32. iloveboys
    • 2 years ago
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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

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

  34. iloveboys
    • 2 years ago
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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

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

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

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

  38. jim_thompson5910
    • 2 years ago
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    you're welcome

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

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

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

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

  43. jim_thompson5910
    • 2 years ago
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    so same name

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

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

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

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

  48. iloveboys
    • 2 years ago
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    ok

  49. jim_thompson5910
    • 2 years ago
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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

  50. iloveboys
    • 2 years ago
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    thank you jim

  51. jim_thompson5910
    • 2 years ago
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    anytime

  52. iloveboys
    • 2 years ago
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    bye

  53. jim_thompson5910
    • 2 years ago
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    cya later

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