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jazzyfa30 Group Title

Solve the equation by using the Square Root Property. x^2 = 10x + 24

  • one year ago
  • one year ago

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  1. jazzyfa30 Group Title
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    @hartnn

    • one year ago
  2. hartnn Group Title
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    ok, so \(x^2-10x=24\) the co-efficient of 'x' here is 10 divide it by 2 and square the result, what u get ?

    • one year ago
  3. hartnn Group Title
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    ***the co-efficient of 'x' here is -10

    • one year ago
  4. jazzyfa30 Group Title
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    so it would be x^2-5x=24

    • one year ago
  5. hartnn Group Title
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    no no.. the co-efficient of x is -10, right ? divide it by 2, you get -5, but then square it, \(\large (-5)^2=...?\)

    • one year ago
  6. jazzyfa30 Group Title
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    2.23

    • one year ago
  7. jazzyfa30 Group Title
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    is that right????

    • one year ago
  8. hartnn Group Title
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    umm...noo...how u got thta ?

    • one year ago
  9. jazzyfa30 Group Title
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    i squared 5 isnt that what i was supposed to do

    • one year ago
  10. hartnn Group Title
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    5^2 =25

    • one year ago
  11. jazzyfa30 Group Title
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    oh i thought i was supposed to square root it not square my b now what

    • one year ago
  12. hartnn Group Title
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    ohh...u took square root...we are supposed to square..

    • one year ago
  13. hartnn Group Title
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    ok, we add 25 on both sides \(x^2-10x+25=24+25\) notice that left side is a perfect square! of ?

    • one year ago
  14. jazzyfa30 Group Title
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    49

    • one year ago
  15. hartnn Group Title
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    49 is right side, and is the square of 7 what about left side ? x^2-10x+25 is squaree of ?

    • one year ago
  16. jazzyfa30 Group Title
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    idk

    • one year ago
  17. hartnn Group Title
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    \((x-5)^2 =... ?\) can u find ?

    • one year ago
  18. jazzyfa30 Group Title
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    x^2- 25

    • one year ago
  19. hartnn Group Title
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    nopes, \((a+b)^2=a^2+2ab+b^2\) so, \((x-5)^2 =x^2-10x+25\) so here we have \((x-5)^2=7^2\) take square root on both sides!

    • one year ago
  20. jazzyfa30 Group Title
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    ok

    • one year ago
  21. jazzyfa30 Group Title
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    then what

    • one year ago
  22. jazzyfa30 Group Title
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    im confused

    • one year ago
  23. hartnn Group Title
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    did u take square root on both sides ? what u got ?

    • one year ago
  24. jazzyfa30 Group Title
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    idk

    • one year ago
  25. hartnn Group Title
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    \((x-5)^2=7^2 \implies x-5 = \pm 7 \\ x-5 =7 \quad \quad x-5=-7\) get 2 values of 'x' from here!

    • one year ago
  26. jazzyfa30 Group Title
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    but 7 isnt apart of any of my answer choices

    • one year ago
  27. hartnn Group Title
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    u need to solve x-5 =7 x-5 = -7

    • one year ago
  28. jazzyfa30 Group Title
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    x= 12 x=-2

    • one year ago
  29. hartnn Group Title
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    correct!

    • one year ago
  30. jazzyfa30 Group Title
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    ok Solve the equation by completing the square: x2 - 8x + 15 = 0

    • one year ago
  31. hartnn Group Title
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    same thing, here, the co-efficient of x is -8 divide it by 2, what u get ?

    • one year ago
  32. jazzyfa30 Group Title
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    -4

    • one year ago
  33. hartnn Group Title
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    yes, then square it...what u get ?

    • one year ago
  34. jazzyfa30 Group Title
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    -16

    • one year ago
  35. hartnn Group Title
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    -4 *-4 will be +16

    • one year ago
  36. jazzyfa30 Group Title
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    oh

    • one year ago
  37. hartnn Group Title
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    so, adding 16 on both sides... \(x^2-8x+16=-15+16\)

    • one year ago
  38. jazzyfa30 Group Title
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    right side:1 left side: idk

    • one year ago
  39. jazzyfa30 Group Title
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    i think its x^2-4x-2x=1

    • one year ago
  40. jazzyfa30 Group Title
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    is that right??????

    • one year ago
  41. jazzyfa30 Group Title
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    @GoldPhenoix please help

    • one year ago
  42. hartnn Group Title
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    \(x^2-8x+16 = (x-4)^2\) so, we have \((x-4)^2=1\) just take square root on both sides!

    • one year ago
  43. jazzyfa30 Group Title
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    ?????????

    • one year ago
  44. hartnn Group Title
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    when you take square root, the square part goes away! (x+4) =+1 or x-4 =-1 find 2 values of x from here...

    • one year ago
  45. jazzyfa30 Group Title
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    (x-3)(x+3) i think

    • one year ago
  46. jazzyfa30 Group Title
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    a. x = + 5 b. x =-3 or x = 5 c. x =3 or x = -5 d. x =3 or x =5

    • one year ago
  47. hartnn Group Title
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    x-4 = -1 , x = -1+4 = 3 is correct but for 2nd part, you get x-4 =1 x= 1+4 = 5

    • one year ago
  48. hartnn Group Title
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    so, 3,5

    • one year ago
  49. jazzyfa30 Group Title
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    oh yea

    • one year ago
  50. jazzyfa30 Group Title
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    Solve the equation by completing the square: x^2 + 2x = 120

    • one year ago
  51. terenzreignz Group Title
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    Do you know how to complete the square? :)

    • one year ago
  52. jazzyfa30 Group Title
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    ??????

    • one year ago
  53. terenzreignz Group Title
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    I see. Here's an example:\[\Large x^2 + \color{red}6x =40 \]

    • one year ago
  54. terenzreignz Group Title
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    What you do is take that coefficient (number beside) the x with NO EXPONENT. In our case, 6.

    • one year ago
  55. terenzreignz Group Title
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    Take half of it, 3, and then square it, getting 9.

    • one year ago
  56. terenzreignz Group Title
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    Now, add that to BOTH SIDES of the equation. \[\Large x^2 +\color{red}6x\color{blue}{+9}=40\color{blue}{+9}\]

    • one year ago
  57. terenzreignz Group Title
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    Simplify. \[\Large x^2+\color{red}6x+9=49\]

    • one year ago
  58. terenzreignz Group Title
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    You'll notice that the left side is now what you call a 'perfect square trinomial' It may be expressed as... \[\Large (x+\color{red}3)^2=49\]

    • one year ago
  59. terenzreignz Group Title
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    Notice of course, that 3 is half of the original coefficient of x, which was 6. Now take the square root of both sides: \[\Large \sqrt{(x+3)^2}=\sqrt{49}\] Notice that the square root and the square (exponent 2) cancel out. But the square root of 49 may be either positive or negative 7. \[\Large x+3=\pm7\] Bring the 3 to the other side, you now get \[\Large x = -3\pm7\] which means either x = -3 + 7 = 4 or x = -3 - 7 = -10 Case Closed :)

    • one year ago
  60. terenzreignz Group Title
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    So... go ahead now, get started with your own problem, using the same concepts. I'll walk you through it if you like :)

    • one year ago
  61. jazzyfa30 Group Title
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    yes please

    • one year ago
  62. terenzreignz Group Title
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    By your lead :)

    • one year ago
  63. jazzyfa30 Group Title
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    u on skype or call me at 803-937-6271 cus by u typing it it is confusing m please

    • one year ago
  64. terenzreignz Group Title
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    I don't have skype. come on, we can do this... I'll get you started... \[\Large x^2 +\color{red}2x = 120\]

    • one year ago
  65. jazzyfa30 Group Title
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    then call me i dont understand

    • one year ago
  66. jazzyfa30 Group Title
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    @jessika96

    • one year ago
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