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BloomLocke367

  • one year ago

I only have two left that I'm a little unclear on.

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  1. BloomLocke367
    • one year ago
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    \(x+1-2\sqrt{x+4}=0\)

  2. BloomLocke367
    • one year ago
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    I have to solve that too

  3. BloomLocke367
    • one year ago
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    I know the 4 can come out of the radical

  4. Nnesha
    • one year ago
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    no

  5. Nnesha
    • one year ago
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    it's (x+4)

  6. BloomLocke367
    • one year ago
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    ohhhhhhhhhhhh

  7. Nnesha
    • one year ago
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    you can't separate it :(

  8. BloomLocke367
    • one year ago
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    yeah, when you wrote it that way I noticed.

  9. Nnesha
    • one year ago
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    so move all the terms to the right side instead sqrt{(x+4}

  10. BloomLocke367
    • one year ago
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    Why am I moving them?

  11. Nnesha
    • one year ago
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    you need to get sqrt{x+4} one one side and all other terms to opposite side so then you can take square both sides to cancel out the square root

  12. BloomLocke367
    • one year ago
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    OH THAT MAKES SENSE

  13. BloomLocke367
    • one year ago
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    i have \(\frac{1}{2}x+\frac{1}{2}=\sqrt{x+4}\)

  14. BloomLocke367
    • one year ago
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    right?

  15. Nnesha
    • one year ago
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    looks right

  16. BloomLocke367
    • one year ago
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    so \(x+4=\frac{1}{4}x+\frac{1}{4}\), right?

  17. Nnesha
    • one year ago
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    mhmm

  18. BloomLocke367
    • one year ago
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    so 0.75x=-3.75?

  19. Nnesha
    • one year ago
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    there is an easier way to do it

  20. Nnesha
    • one year ago
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    i hate fractions :P \[\huge\rm x+1-2\sqrt{x+4}=0\] move the x+1 to the right side

  21. BloomLocke367
    • one year ago
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    \(-2\sqrt{x+4}=-x-1\)

  22. Nnesha
    • one year ago
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    looks good now move the -2 to the right side remember we need just radical sign on one side

  23. BloomLocke367
    • one year ago
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    I already did that.

  24. Nnesha
    • one year ago
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    \(\color{Red}{-2}\sqrt{x+4}=-x-1\) we just need radical at left side

  25. BloomLocke367
    • one year ago
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    oops

  26. BloomLocke367
    • one year ago
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    \(\sqrt{x+4}=\frac{1}{2}x+\frac{1}{2}\)

  27. BloomLocke367
    • one year ago
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    but, look:\(\color{#0cbb34}{\text{Originally Posted by}}\) @BloomLocke367 i have \(\frac{1}{2}x+\frac{1}{2}=\sqrt{x+4}\) \(\color{#0cbb34}{\text{End of Quote}}\) I already did that

  28. Nnesha
    • one year ago
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    yes \[\sqrt{x+4}=\frac{ (x+1) }{ 2 }\] which is same as 1/2x+1/2 but keep that to one fraction

  29. Nnesha
    • one year ago
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    when we finish with this question please pnch on my forehead okay

  30. BloomLocke367
    • one year ago
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    hahaha okay XD

  31. Nnesha
    • one year ago
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    that's right but i don't how u got the negative sign

  32. Nnesha
    • one year ago
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    i'm sorry i'm tired actually so i apologies

  33. Nnesha
    • one year ago
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    okay so how would you cancel out the square root ?

  34. BloomLocke367
    • one year ago
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    square both sides

  35. Nnesha
    • one year ago
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    yes right

  36. BloomLocke367
    • one year ago
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    so \(x+4=\frac{1}{4}x+\frac{1}{4}\)

  37. BloomLocke367
    • one year ago
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    which I also said already

  38. Nnesha
    • one year ago
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    \[\sqrt{x+4}= (\frac{x+1 }{ 2 })^2\] that's how you should take square root at left side

  39. BloomLocke367
    • one year ago
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    so \(x+4=\frac{x^2+1}{4}\)

  40. Nnesha
    • one year ago
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    that's why i said it's better to combine them together and no

  41. BloomLocke367
    • one year ago
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    *facepalm*

  42. Nnesha
    • one year ago
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    \[\huge\rm (\frac{ x+1 }{ 2 }) = \frac{ (x+1)^2 }{ 2^2 }\] both should be squared

  43. BloomLocke367
    • one year ago
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    sorry, everyone is talking around me and I just found out my boyfriend is in the ER... I'm a little distracted

  44. Nnesha
    • one year ago
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    mah don't `facepalm` :P

  45. Nnesha
    • one year ago
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    it's okay! :=)

  46. BloomLocke367
    • one year ago
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    oh right, that makes sense

  47. BloomLocke367
    • one year ago
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    x+2x+1 should be the numerator, right?

  48. Nnesha
    • one year ago
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    right now (x+1)^2 is same as (x+1)(x+1) you're an expert when it comes to foil method

  49. Nnesha
    • one year ago
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    typo

  50. Nnesha
    • one year ago
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    you sure it's x+2x +1 ?

  51. BloomLocke367
    • one year ago
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    X^2

  52. BloomLocke367
    • one year ago
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    UGH sorry

  53. Nnesha
    • one year ago
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    yes right :=) \[x+4=\frac{ x^2+2x+1 }{ 4 }\] now oslve for x

  54. Nnesha
    • one year ago
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    solve*

  55. BloomLocke367
    • one year ago
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    4x+16=x^2+2x+1?

  56. BloomLocke367
    • one year ago
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    to start with, that is

  57. Nnesha
    • one year ago
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    :=)

  58. Nnesha
    • one year ago
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    yes that's right

  59. BloomLocke367
    • one year ago
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    okay. so x^2-2x-15?

  60. Nnesha
    • one year ago
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    mhmm error!

  61. BloomLocke367
    • one year ago
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    what?

  62. BloomLocke367
    • one year ago
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    what error?

  63. Nnesha
    • one year ago
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    \[\color{ReD}{4x}+16=x^2\color{ReD}{+}2x+1\]

  64. Nnesha
    • one year ago
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    you would subtract 2x both sides right so 4x-2x = 2x not -2x :=)

  65. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @BloomLocke367 okay. so x^2-2x-15? \(\color{blue}{\text{End of Quote}}\) so it's \[\huge\rm x^2+2x-15=0\]

  66. BloomLocke367
    • one year ago
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    no... 2x-4x=-2x

  67. Nnesha
    • one year ago
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    facedesk**

  68. Nnesha
    • one year ago
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    i moved all the terms to the left side why i'm still looking on it you're right god:(

  69. BloomLocke367
    • one year ago
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    hahaha, we both are having a tough time

  70. BloomLocke367
    • one year ago
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    meanwhile, I'm about to have a meltdown because I'm so worried...

  71. BloomLocke367
    • one year ago
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    anyhoo, I have (x-5)(x+3)=0

  72. Nnesha
    • one year ago
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    ah my fault

  73. BloomLocke367
    • one year ago
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    so my zeros are 5 and -3, right?

  74. Nnesha
    • one year ago
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    yes right solve for x and what is the statement ? you have to find solutions right ?

  75. BloomLocke367
    • one year ago
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    yep

  76. BloomLocke367
    • one year ago
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    which I found

  77. Nnesha
    • one year ago
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    make sure you check your work plugin 5 and -3 into the equation

  78. Nnesha
    • one year ago
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    there can be an extraneous solution!

  79. BloomLocke367
    • one year ago
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    yea, I know. I just have one more.. then I can go and quietly breakdown in my room. ;-;

  80. Nnesha
    • one year ago
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    i'm pretty you don't want to help anymore ahahhah

  81. BloomLocke367
    • one year ago
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    5 works. just checked

  82. Nnesha
    • one year ago
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    me*

  83. Nnesha
    • one year ago
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    yes right -3 is an extraneous

  84. BloomLocke367
    • one year ago
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    yep

  85. Nnesha
    • one year ago
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    alright good luck btw what's ur next question ?

  86. BloomLocke367
    • one year ago
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    \(\sqrt x+x=1\)

  87. Nnesha
    • one year ago
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    is it okay if we do it on this thread ??

  88. BloomLocke367
    • one year ago
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    yea, I don't mind. I just wanna get this done

  89. Nnesha
    • one year ago
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    okay cool! that's the easy one just like we did move the x to the right side (bec we need the radical one side )

  90. BloomLocke367
    • one year ago
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    \(\sqrt x=-x+1\)

  91. Nnesha
    • one year ago
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    yes right take square both sides \[(\sqrt{x})^2= (-x+1)^2\]

  92. Nnesha
    • one year ago
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    (-x+1)(-x+1) =foil

  93. BloomLocke367
    • one year ago
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    x=x^2-2x+1

  94. Nnesha
    • one year ago
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    right solve for x :=)

  95. BloomLocke367
    • one year ago
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    I'm drawing a blank here, you know why. should I factor the right side, or move the x on the left to the right?

  96. Nnesha
    • one year ago
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    well should move the x to the right side cuz there are like terms that we can combine

  97. BloomLocke367
    • one year ago
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    ok

  98. Nnesha
    • one year ago
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    if you factor right side that will not help you you still have to distribute factors with x

  99. BloomLocke367
    • one year ago
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    x^2-3x+1

  100. Nnesha
    • one year ago
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    yes right you need to apply quadratic formula to find x

  101. BloomLocke367
    • one year ago
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    ok

  102. BloomLocke367
    • one year ago
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    I can do that

  103. BloomLocke367
    • one year ago
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    gimme a moment

  104. Nnesha
    • one year ago
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    cuz one is the only one factor of 1 so 1 time 1 there is not way you can get 3 from 1+1

  105. Nnesha
    • one year ago
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    take your time! :=)

  106. BloomLocke367
    • one year ago
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    yea I know gimme a sec

  107. BloomLocke367
    • one year ago
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    so far I have \(x=\frac{3\pm\sqrt5}{2}\)

  108. BloomLocke367
    • one year ago
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    that's as far as I'm going

  109. BloomLocke367
    • one year ago
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    is that right?

  110. BloomLocke367
    • one year ago
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    @Nnesha

  111. Nnesha
    • one year ago
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    looks right!

  112. BloomLocke367
    • one year ago
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    ok. I'm leaving it as is

  113. BloomLocke367
    • one year ago
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    I'm gonna go now, thanks for your help. Goodbye!

  114. Nnesha
    • one year ago
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    alright good luck!

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