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some.random.cool.kid

  • one year ago

Solve the equation 4x2 − 12x + 3 = 0 by completing the square. Verify your answer using the quadratic formula.

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  1. some.random.cool.kid
    • one year ago
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    @Nnesha

  2. some.random.cool.kid
    • one year ago
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    @nincompoop

  3. some.random.cool.kid
    • one year ago
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    i have a few of these this is just the first that i need help with...

  4. some.random.cool.kid
    • one year ago
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    @ParthKohli

  5. some.random.cool.kid
    • one year ago
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    @peachpi

  6. some.random.cool.kid
    • one year ago
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    @whpalmer4

  7. some.random.cool.kid
    • one year ago
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    ik the forumla is this: x = [ -b ± sqrt(b^2 - 4ac) ] / 2a

  8. Nnesha
    • one year ago
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    well we should use complete the square method you can move the constant term to the right side

  9. some.random.cool.kid
    • one year ago
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    x is the term correct?

  10. Nnesha
    • one year ago
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    \[ \huge\rm 4x^2 − 12x + 3 = 0\] subtract 3 both sides

  11. Nnesha
    • one year ago
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    not just x term are divided by plus or negative sign

  12. some.random.cool.kid
    • one year ago
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    uh... subtract the 4x^2 by 3 and the 12 by 3?

  13. Nnesha
    • one year ago
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    \[\huge\rm Ax^2+Bx+C=0\]quadratic e quation where A=leading coefficient B=middle term C=constant term(number with no variable )

  14. Nnesha
    • one year ago
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    |dw:1443453772906:dw| you can't combine -3 and 4x^2 they arn't like terms

  15. some.random.cool.kid
    • one year ago
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    im not sure what you do next after you get -3..

  16. some.random.cool.kid
    • one year ago
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    divide?

  17. Nnesha
    • one year ago
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    |dw:1443453908051:dw| we should complete the square of the left part

  18. Nnesha
    • one year ago
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    \[\huge\rm \color{Red}{4x^2+12x}=-3\] what is the common factor at left side ?

  19. some.random.cool.kid
    • one year ago
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    the x?

  20. Nnesha
    • one year ago
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    no just take out the common number not the varaible

  21. some.random.cool.kid
    • one year ago
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    uh ... im confused. am i supposed to take out the 4 and 12 and divide them?

  22. some.random.cool.kid
    • one year ago
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    besides the x i see nothing in common...

  23. Nnesha
    • one year ago
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    well it's 4x^2+12x so we just need to take out the common number

  24. Nnesha
    • one year ago
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    what are the factors of 4 and 12 ?

  25. some.random.cool.kid
    • one year ago
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    1, 2, 4 ... 4 is best factor... then i divide both by four to get 3 right?

  26. Nnesha
    • one year ago
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    you need GREATEST COMMON FACTOR 1,2,4 are factors of 4 1,2,3,4,6,12 are factors of 12 so which number is common ?

  27. Nnesha
    • one year ago
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    greatest number **

  28. some.random.cool.kid
    • one year ago
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    uh i said 4...

  29. some.random.cool.kid
    • one year ago
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    is there one greater than thta?

  30. some.random.cool.kid
    • one year ago
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    that*

  31. Nnesha
    • one year ago
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    i don't understand what u mean "divide `both` to get 3"

  32. Nnesha
    • one year ago
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    \[\huge\rm 4(x^2+3x)=-3\]when you take out 4 from 4 u will get 1

  33. some.random.cool.kid
    • one year ago
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    you still have the 12x

  34. some.random.cool.kid
    • one year ago
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    i was saying didvide the 4x^2 and 12x by the common factor to simplify the solution.

  35. Nnesha
    • one year ago
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    now we can complete the square of (x^2+3x) we should divide b by b here is an example \[\huge\rm a(x^2+bx)=-C\] divide b by 2 \[\huge\rm a(x+\frac{ b }{ 2 })^2=-C + (\frac{ b }{ 2})^2 \] and then add(b/2)^2 to both sides

  36. Nnesha
    • one year ago
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    yes right. i read it wrong. so sorry

  37. Nnesha
    • one year ago
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    \[\huge\rm 4(x^2+3x)=-3\] take half of the x term and then square it add square to the both sides

  38. some.random.cool.kid
    • one year ago
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    you have to x's which one do i take half of?

  39. some.random.cool.kid
    • one year ago
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    two*

  40. Nnesha
    • one year ago
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    first one is x^2 2nd one is x

  41. some.random.cool.kid
    • one year ago
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    so take half of the second one?

  42. Nnesha
    • one year ago
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    yes right

  43. some.random.cool.kid
    • one year ago
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    uh... ok so now you'll have- x+ 3?

  44. some.random.cool.kid
    • one year ago
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    thats not a negative sign

  45. Nnesha
    • one year ago
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    x+3 how ? we should take half of 3 in other words just divide `b by 2`

  46. some.random.cool.kid
    • one year ago
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    how am i dividing 3x by 2?

  47. some.random.cool.kid
    • one year ago
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    /i thought you wanted me to take away x on both sides..

  48. Nnesha
    • one year ago
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    here is an example \[\huge\rm a(x^2+bx)=-C\] divide b by 2 \[\huge\rm a(x+\frac{ b }{ 2 })^2=-C + (\frac{ b }{ 2})^2 \times a \] and then add(b/2)^2 to both sides here is an example

  49. Nnesha
    • one year ago
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    well first we need to divide b by 2 take square of (b/2)^2 then add (b/2)^2 to both sides

  50. Nnesha
    • one year ago
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    what is b in ur equation ?

  51. some.random.cool.kid
    • one year ago
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    the 3x?

  52. Nnesha
    • one year ago
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    just b

  53. Nnesha
    • one year ago
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    3 =b not 3x x is variable

  54. Nnesha
    • one year ago
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    yes right \[\huge\rm 4(x^2+\color{blue}{\frac{ 3x }{ 2 }}+\color{Red}{(\frac{ 3 }{ 2 })^2}) = -3 +\color{red}{(\frac{3}{2})^2 } \times 4\] look at the blue term i took half of 3 and then add (b/2)^2 to both sides and sicne we got the common factor we should multiply (b/2)^2 by 4 at right side

  55. some.random.cool.kid
    • one year ago
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    do we get rid of negative 3 with the positive 3

  56. Nnesha
    • one year ago
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    positive 3 ?? it's 3x you can't combine 3x with -3 they are not like terms

  57. some.random.cool.kid
    • one year ago
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    then your left with whats in blue right?

  58. Nnesha
    • one year ago
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    right we can factor this quadraticc equation (x^2 +3/2x + (3/2)^2) or look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\]

  59. Nnesha
    • one year ago
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    you need to convert left side in squared form

  60. some.random.cool.kid
    • one year ago
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    so its a bit of a leap here but i believe the simplified form would be -3 +- 2\[\sqrt{6}\] over 2?

  61. some.random.cool.kid
    • one year ago
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    everything over 2*

  62. Nnesha
    • one year ago
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    and how did you get that ?

  63. some.random.cool.kid
    • one year ago
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    well you told me that like terms cant combine

  64. some.random.cool.kid
    • one year ago
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    leaving negative 3 alone.

  65. Nnesha
    • one year ago
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    look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\] and tell me how would you write \[(x^2+3x) = ????\]

  66. some.random.cool.kid
    • one year ago
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    then you said something about b being divided by2? which i believe gave me that..

  67. Nnesha
    • one year ago
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    how did you write \[(x^2+3x)\] in squared form ?

  68. some.random.cool.kid
    • one year ago
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    mmm

  69. some.random.cool.kid
    • one year ago
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    3x^2? idk

  70. some.random.cool.kid
    • one year ago
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    ik your left with one x right?

  71. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha look at this example \[\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2\] and tell me how would you write \[(x^2+3x) = ????\] \(\color{blue}{\text{End of Quote}}\)

  72. some.random.cool.kid
    • one year ago
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    thats where im lost though...

  73. Nnesha
    • one year ago
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    yea that's obvious we have to solve for x

  74. Nnesha
    • one year ago
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    then how you got the final x value ?

  75. some.random.cool.kid
    • one year ago
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    i got what?

  76. Nnesha
    • one year ago
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    here is another example \[(x^2+6x)=(x+\frac{ 6 }{ 2})^2=(x+3)^2\]

  77. some.random.cool.kid
    • one year ago
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    wait but isn't that similar to what i had?

  78. Nnesha
    • one year ago
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    similar to what ?

  79. Nnesha
    • one year ago
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    \[\huge\rm 4(x^2+\color{blue}{\frac{ 3x }{ 2 }}+\color{Red}{(\frac{ 3 }{ 2 })^2}) = -3 +\color{red}{(\frac{3}{2})^2 } \times 4\] \[4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4\] according to what i said (x^2+bx)=(x+b/2)^2

  80. Nnesha
    • one year ago
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    solve this to get the answer \[4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4\]

  81. Nnesha
    • one year ago
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    here is how we can write in squared form \[(x^2+bx)=(x+\frac{ b }{ 2 })^2\] \[(x^2+6x)=x^2+6x+\color{reD}{9}\] i divided 6 by 2 which is equal to 3 then squared 3 that's how i got 9 now we can factor x^2+6x+9 (x+3)(x+3) are factors of that quadratic equation which is same as (x+3)^2 so in short we can do this \[(x^2+bx)=(x+\frac{ b }{ 2})^2\]\[(x^2+6x)=(x+\frac{ 6 }{ 2 })^2=(x+3)^2\]

  82. Nnesha
    • one year ago
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    feel free to ask question about that^let me know if you have question on that

  83. some.random.cool.kid
    • one year ago
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    ok thank you

  84. some.random.cool.kid
    • one year ago
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    i had another one mind helping?

  85. some.random.cool.kid
    • one year ago
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    @Nnesha

  86. Nnesha
    • one year ago
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    did you get this one ?

  87. some.random.cool.kid
    • one year ago
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    i believe i did

  88. some.random.cool.kid
    • one year ago
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    Solve x2 − 5x + 1 = 0 using the quadratic formula.

  89. Nnesha
    • one year ago
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    alright i'll try

  90. some.random.cool.kid
    • one year ago
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    i know the first step to this ...

  91. Nnesha
    • one year ago
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    can u make a new post bec this one is laggy

  92. some.random.cool.kid
    • one year ago
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    i have noticed

  93. Nnesha
    • one year ago
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    ye it keeps moving hard to use latex on this post

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