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anonymous

  • one year ago

Use the quadratic formula to solve 2y2 – 3y = 2. A. {½, 2} B. {½, –2} C. {–½, –2} D. {–½, 2}

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  1. pooja195
    • one year ago
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    Before the subtract the 2 so the the equation is = to 0 \[\huge 2y^2-3y-2\]

  2. pooja195
    • one year ago
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    What are your a b c values?

  3. pooja195
    • one year ago
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    Once you have those we can plug it into this equation AKA Quadratic Formula

  4. pooja195
    • one year ago
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    \[\huge~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

  5. anonymous
    • one year ago
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    I barley started to learn this, do you think you can walk me through it, step by step?

  6. pooja195
    • one year ago
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    \[\huge~ax^2+bx+c\] \[\huge~2y^2-3y-2=0\] Anything that is ^2 will be the A just tell the number ignore the y^2 then anything with a plain vairble will just be the b value in this case it is -3 A number alone is the C value which in this case is -2

  7. pooja195
    • one year ago
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    Got it so far?

  8. anonymous
    • one year ago
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    yes

  9. anonymous
    • one year ago
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    @pooja195

  10. pooja195
    • one year ago
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    Good now we plug it in

  11. KendrickLamar2014
    • one year ago
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    Step 1. Subtract 2 from both sides: \[2y^2−3y−2=2−2\] \[= 2y^2−3y−2=0\]

  12. anonymous
    • one year ago
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    okay! @KendrickLamar2014

  13. KendrickLamar2014
    • one year ago
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    Step 2. Factor the Left side: \[(2y+1)(y−2)=0\]

  14. pooja195
    • one year ago
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    \[\huge~x=\frac{ -(-3)\pm \sqrt{(-3)^2-4(2)(-2)} }{ 2(2) } \]

  15. KendrickLamar2014
    • one year ago
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    Step 3. Set the factors to equal 0: \[2y+1=0~ or ~y−2=0\]

  16. pooja195
    • one year ago
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    @KendrickLamar2014 it says to use the quadratic formula not factoring

  17. KendrickLamar2014
    • one year ago
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    I know, but you can get the same answer.

  18. pooja195
    • one year ago
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    They need to use the quadratic formula its the directions.

  19. KendrickLamar2014
    • one year ago
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    \[= y = -\frac{ 1 }{ 2 }~or~ y = 2\]

  20. pooja195
    • one year ago
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    Or just give them the answer...... @ninowletonzalez06 are you allowed to use factoring?

  21. KendrickLamar2014
    • one year ago
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    Yah, I was just showing her that there are many ways to solve this equation

  22. pooja195
    • one year ago
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    Or have u not learned it because the Quadratic formula comes before factoring @KendrickLamar2014

  23. pooja195
    • one year ago
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    Well now i bet that the asker is confused .-.

  24. anonymous
    • one year ago
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    a liitle, can you explain it in the easiest form!

  25. pooja195
    • one year ago
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    do you HAVE to use the quadratic formula?

  26. anonymous
    • one year ago
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    by what the question is asking yes!

  27. anonymous
    • one year ago
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    are you still there?

  28. pooja195
    • one year ago
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    Ok :) All i did here was plug in the values

  29. pooja195
    • one year ago
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    LaTex isnt working i cant finish it .-.

  30. pooja195
    • one year ago
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    \[\huge~x=\frac{ -(-3) \pm \sqrt{(-3)^2-4(2)(-2)} }{ 2(2)} \]

  31. anonymous
    • one year ago
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    okay!:)

  32. pooja195
    • one year ago
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    \[\huge~x=\frac{ 3 \pm \sqrt{(-3)^2-4(2)(-2)} }{ 2(2)} \] simplify whats in the square root but DO NOT square root it

  33. anonymous
    • one year ago
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    |dw:1433005023557:dw|

  34. anonymous
    • one year ago
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    @pooja195

  35. pooja195
    • one year ago
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    (-3)^2-4(2)(-2) thats all just miltiply :P

  36. pooja195
    • one year ago
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    *multiply

  37. anonymous
    • one year ago
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    9-(-16)

  38. pooja195
    • one year ago
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    hmmm nope let me help u out :) i plugged it into a calc and got 25 \[\huge~x=\frac{ 3 \pm \sqrt{25} }{ 4} \]

  39. anonymous
    • one year ago
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    oh okay!:)

  40. pooja195
    • one year ago
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    Now we turn it into 2 diffrent problems \[\huge \frac{ 3+\sqrt{25} }{ 4 }\] and \[\huge \frac{ 3-\sqrt{25} }{ 4 }\]

  41. pooja195
    • one year ago
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    Do you know how to solve?

  42. anonymous
    • one year ago
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    3-5/4

  43. anonymous
    • one year ago
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    @pooja195

  44. pooja195
    • one year ago
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    im kinda confused on the solvng part :/ ive always used a clac to solve @KendrickLamar2014 do you know?

  45. anonymous
    • one year ago
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    help!!!

  46. pooja195
    • one year ago
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    @ybarrap can you help solve .-. please :)

  47. KendrickLamar2014
    • one year ago
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    What?

  48. pooja195
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @pooja195 Now we turn it into 2 diffrent problems \[\huge \frac{ 3+\sqrt{25} }{ 4 }\] and \[\huge \frac{ 3-\sqrt{25} }{ 4 }\] \(\color{blue}{\text{End of Quote}}\) this

  49. pooja195
    • one year ago
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    i use a calc to solve that last step i dont know how to explain it

  50. KendrickLamar2014
    • one year ago
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    Oh, to explain or to solve it?

  51. anonymous
    • one year ago
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    solve it

  52. ybarrap
    • one year ago
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    @pooja195 looks like you've got it covered here

  53. KendrickLamar2014
    • one year ago
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    I belive you take the square root of 25, then you add/subtract 3, lastly divide by 4

  54. KendrickLamar2014
    • one year ago
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    Ye, thats how you solve it ^

  55. anonymous
    • one year ago
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    do u add or subtract

  56. KendrickLamar2014
    • one year ago
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    both

  57. KendrickLamar2014
    • one year ago
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    Its two equations

  58. anonymous
    • one year ago
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    i confused

  59. anonymous
    • one year ago
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    im**

  60. KendrickLamar2014
    • one year ago
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    \[\frac{ 3 + \sqrt25 }{ 4 }\] \[\frac{ 3-\sqrt25 }{ 4 }\] Two equations

  61. KendrickLamar2014
    • one year ago
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    \[\sqrt 25 = 5\] \[3+5 = 8\]

  62. anonymous
    • one year ago
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    okay! i understand now!

  63. KendrickLamar2014
    • one year ago
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    \[\frac{ 8 }{ 4 } = 2\]

  64. anonymous
    • one year ago
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    yes

  65. KendrickLamar2014
    • one year ago
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    Can you solve the other equation?

  66. anonymous
    • one year ago
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    2/4=2

  67. KendrickLamar2014
    • one year ago
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    This equation:\[\frac{ 3-\sqrt25 }{ 4}\]

  68. anonymous
    • one year ago
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    so -2?

  69. KendrickLamar2014
    • one year ago
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    No, lets go step-by-step. What is the square root of 25?

  70. anonymous
    • one year ago
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    5

  71. KendrickLamar2014
    • one year ago
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    Yes, now what is 3-5?

  72. anonymous
    • one year ago
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    -2

  73. KendrickLamar2014
    • one year ago
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    Yes, now what is \[\frac{ -2 }{ 4 }\]

  74. anonymous
    • one year ago
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    -1/2

  75. anonymous
    • one year ago
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    so the answer is D?

  76. KendrickLamar2014
    • one year ago
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    Yes

  77. KendrickLamar2014
    • one year ago
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    D is Correct

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