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P.nut1996

  • 3 years ago

Please help me solve? I'm looking at this problem and not seeing how to work it so that my answer comes out looking like the choices...

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  1. P.nut1996
    • 3 years ago
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    \[2\sqrt{44x ^{3}}-\sqrt{7}-\sqrt{99x ^{3}}+\sqrt{63}\]

  2. P.nut1996
    • 3 years ago
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    Combine like terms, right? I don't see how to combine \[2\sqrt{44x ^{3}}\] and \[-\sqrt{99x ^{3}}\]

  3. javawarrior
    • 3 years ago
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    before combining see if you can simplify each term first

  4. P.nut1996
    • 3 years ago
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    How so...?

  5. P.nut1996
    • 3 years ago
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    Oh! like 11 goes into both 44 and 99??

  6. javawarrior
    • 3 years ago
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    see if u can get rid of the roots!

  7. P.nut1996
    • 3 years ago
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    How would I go about doing that? Right now I have \[2\sqrt{4x ^{3}}-\sqrt{7}-\sqrt{9x ^{3}}+\sqrt{63}\]

  8. P.nut1996
    • 3 years ago
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    Does it look right? Going in the right direction maybe?

  9. P.nut1996
    • 3 years ago
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    Oh, and I eliminated 7. and the sqrt of 63 is now the sqrt of 9

  10. hartnn
    • 3 years ago
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    \(\sqrt {63} = \sqrt {9} \sqrt {7}=...?\)

  11. P.nut1996
    • 3 years ago
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    Now it looks like \[2\sqrt{4x ^{3}}-\sqrt{9x ^{3}}+\sqrt{9}\]

  12. hartnn
    • 3 years ago
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    \(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...?\) you just can't eliminate 7...

  13. Andresfon12
    • 3 years ago
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    something is missing

  14. hartnn
    • 3 years ago
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    \(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}=...? \\\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?\)

  15. P.nut1996
    • 3 years ago
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    But & goes into 7 one time and it goes into 63 9 times. Simplifying... Right?

  16. P.nut1996
    • 3 years ago
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    but 7*

  17. hartnn
    • 3 years ago
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    so, \(\sqrt {63}-\sqrt 7 = \sqrt{9}\sqrt{7}-\sqrt{7} = \sqrt{7}[3-1]=..?\)

  18. hartnn
    • 3 years ago
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    you see what i did there ^ ?

  19. P.nut1996
    • 3 years ago
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    No... :(

  20. Andresfon12
    • 3 years ago
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    @hartnn i see what he did

  21. Andresfon12
    • 3 years ago
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    3x3

  22. hartnn
    • 3 years ago
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    \(\sqrt {xy}=\sqrt{x}\sqrt{y}\) ok ? so, \(\sqrt{63}=\sqrt{9}\sqrt{7}=3 \sqrt{7}\) right ?

  23. hartnn
    • 3 years ago
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    ask if any doubts.

  24. P.nut1996
    • 3 years ago
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    That does not look like any of my answer choices :(

  25. hartnn
    • 3 years ago
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    that was just simplification of constant terms.... \(3\sqrt 7- \sqrt 7= 2 \sqrt 7\) simplification of 'x' terms is still remaining. and more important thing is that you understand....so that you can do other problems on your own...

  26. P.nut1996
    • 3 years ago
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    I have no idea.

  27. hartnn
    • 3 years ago
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    so, do you want me to start over ? or take it from a particular step ?

  28. P.nut1996
    • 3 years ago
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    Start from the beginning please?

  29. hartnn
    • 3 years ago
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    right, lets see term by term, so 63 is 9 times 7 so, \(\sqrt {63}= \sqrt {9}\sqrt{7}=...?\)

  30. P.nut1996
    • 3 years ago
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    Yes. And what do we do with the \[\sqrt{7}\] just leave it where it is?

  31. hartnn
    • 3 years ago
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    i was getting there... you first tell me what will be \(\sqrt 9 \sqrt7 =... ?\)

  32. P.nut1996
    • 3 years ago
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    simplify by dividing both \[\sqrt{7}\] and \[\sqrt{63}\] by 7?

  33. hartnn
    • 3 years ago
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    dividing ? why ? no..

  34. P.nut1996
    • 3 years ago
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    Oh... Grrr

  35. hartnn
    • 3 years ago
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    lets finish 1 term entirely \(\sqrt {63}=\sqrt 9 \sqrt7 =... ?\)

  36. P.nut1996
    • 3 years ago
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    Yeah. 3.

  37. hartnn
    • 3 years ago
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    so, \(\sqrt {63}=\sqrt 9 \sqrt7 =... ?\)

  38. P.nut1996
    • 3 years ago
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    \[\sqrt{3}\] \[\sqrt{2.6}??\]

  39. hartnn
    • 3 years ago
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    since \(\sqrt 9=3\) \(\sqrt {63}=\sqrt 9 \sqrt7 =3 \sqrt 7\) got this ?

  40. mathstudent55
    • 3 years ago
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    |dw:1359950980241:dw|

  41. P.nut1996
    • 3 years ago
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    \[3\sqrt{7}\]

  42. mathstudent55
    • 3 years ago
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    |dw:1359951141309:dw|

  43. hartnn
    • 3 years ago
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    ok, so now you have constant terms as \(3\sqrt 7-\sqrt 7\) can you combine this ? by factoring out \(\sqrt 7 \)

  44. hartnn
    • 3 years ago
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    \(3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=...?\) ok with this ?

  45. P.nut1996
    • 3 years ago
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    Wait...at the end multiplying \[\sqrt{7}\] by 2?

  46. mathstudent55
    • 3 years ago
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    |dw:1359951264530:dw|

  47. hartnn
    • 3 years ago
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    \(3\sqrt 7-\sqrt 7 = \sqrt 7 [3-1]=2 \sqrt 7\)

  48. P.nut1996
    • 3 years ago
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    So now we have part of it completed?

  49. hartnn
    • 3 years ago
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    yes, now take the terms with 'x'.

  50. P.nut1996
    • 3 years ago
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    11 got into both

  51. P.nut1996
    • 3 years ago
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    goes*

  52. hartnn
    • 3 years ago
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    so, did you get this ? \(\sqrt {44x^3} = \sqrt {4} \sqrt{x^2} \sqrt{11x}\)

  53. hartnn
    • 3 years ago
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    44 = 4*11 x^3 = x*x^2

  54. P.nut1996
    • 3 years ago
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    Why is there no 2?

  55. P.nut1996
    • 3 years ago
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    in front of the \[\sqrt{44x^3}\]

  56. hartnn
    • 3 years ago
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    there is 2 .. \(2\sqrt {44x^3} = 2\sqrt {4} \sqrt{x^2} \sqrt{11x}=2 \times 2x \sqrt {11x}= 4x \sqrt{11x}\)

  57. P.nut1996
    • 3 years ago
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    Okay :) \[4x \sqrt{11x}\]

  58. hartnn
    • 3 years ago
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    similarly, what will be \(\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=...?\)

  59. P.nut1996
    • 3 years ago
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    Uh...?

  60. hartnn
    • 3 years ago
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    \(\sqrt {99x^3} = \sqrt {9} \sqrt{x^2} \sqrt{11x}=3x\sqrt {11x}\) because \(\sqrt 9=3, \: and \: \sqrt {x^2}=x\)

  61. P.nut1996
    • 3 years ago
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    Oh, I wasn't sure how to break it down. I see what you did now :)

  62. P.nut1996
    • 3 years ago
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    \[x \sqrt{11x}+2\sqrt{7}\] :D

  63. hartnn
    • 3 years ago
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    yes. thats correct . i hope you got each step.

  64. P.nut1996
    • 3 years ago
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    Yes! :D Haha thanks

  65. hartnn
    • 3 years ago
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    welcome ^_^

  66. Andresfon12
    • 3 years ago
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    this took a lot time :P

  67. P.nut1996
    • 3 years ago
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    I'm slow, sorry xP

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