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jazzyfa30

  • one year ago

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

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  1. jazzyfa30
    • one year ago
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    @hartnn

  2. hartnn
    • one year ago
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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 ?

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

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

  5. hartnn
    • one year ago
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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, \(\large (-5)^2=...?\)

  6. jazzyfa30
    • one year ago
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    2.23

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

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

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

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

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

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

  13. hartnn
    • one year ago
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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 ?

  14. jazzyfa30
    • one year ago
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    49

  15. hartnn
    • one year ago
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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 ?

  16. jazzyfa30
    • one year ago
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    idk

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

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

  19. hartnn
    • one year ago
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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!

  20. jazzyfa30
    • one year ago
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    ok

  21. jazzyfa30
    • one year ago
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    then what

  22. jazzyfa30
    • one year ago
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    im confused

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

  24. jazzyfa30
    • one year ago
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    idk

  25. hartnn
    • one year ago
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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!

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

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

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

  29. hartnn
    • one year ago
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    correct!

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

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

  32. jazzyfa30
    • one year ago
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    -4

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

  34. jazzyfa30
    • one year ago
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    -16

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

  36. jazzyfa30
    • one year ago
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    oh

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

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

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

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

  41. jazzyfa30
    • one year ago
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    @GoldPhenoix please help

  42. hartnn
    • one year ago
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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!

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

  44. hartnn
    • one year ago
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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...

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

  46. jazzyfa30
    • one year ago
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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

  47. hartnn
    • one year ago
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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

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

  49. jazzyfa30
    • one year ago
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    oh yea

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

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

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

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

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

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

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

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

  58. terenzreignz
    • one year ago
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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... \[\Large (x+\color{red}3)^2=49\]

  59. terenzreignz
    • one year ago
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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 :)

  60. terenzreignz
    • one year ago
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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 :)

  61. jazzyfa30
    • one year ago
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    yes please

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

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

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

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

  66. jazzyfa30
    • one year ago
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    @jessika96

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