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kathyworld09

1.) [4^3 sqrt. a] + [^3 sqrt. 64a] 2.) [sqrt. 5] - [7 sqrt. 80] + [4 sqrt. 45]

  • 3 years ago
  • 3 years ago

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  1. siddharth_tiwari
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    u guyz ought to help everyone.......plz help me also.......

    • 3 years ago
  2. polpak
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    If you can convert the bit under the radical into a product of some squares you can pull out the square and you might end up with something you can combine with addition/subtraction.

    • 3 years ago
  3. polpak
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    Sidd, please stop stalking me.

    • 3 years ago
  4. a933793
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    2) -15 Sqrt[5] not exactly what the first is asking though

    • 3 years ago
  5. kathyworld09
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    siddharth tiwari : ....... well i need help with these so ? ......

    • 3 years ago
  6. polpak
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    a933793: please don't just give people answers. If they want the solution there's wolfram alpha. If they want to learn it'll take more explanation than what you've provided.

    • 3 years ago
  7. siddharth_tiwari
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    k..........just got anxious...u kno....when u dont have any way out to do the quesn ,,,,how does it feels.......well dont mind sir....was just a request not an order.....sry if it mean here

    • 3 years ago
  8. polpak
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    Kathy, can you factor out some of the product under the sqrt and make it into a product (multiplication) of some squared factors?

    • 3 years ago
  9. kathyworld09
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    i have no idea polopak , i get so confused with these sqrts :/

    • 3 years ago
  10. polpak
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    Don't worry about the square roots for now. Can you factor 80?

    • 3 years ago
  11. kathyworld09
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    well the common gcf of all 5, 80, 45 is 5

    • 3 years ago
  12. kathyworld09
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    for problem 2

    • 3 years ago
  13. polpak
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    Not greatest common factor, Just rewrite them as a product of their factors. \(80 = 8 *10 = (4*2) * (5*2) \) \(= (2*2*2)*(5*2) = (2*2)*(2*2) * 5\) \(= 2^2 * 2^2 * 5\)

    • 3 years ago
  14. polpak
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    Now try doing the same thing for 45.

    • 3 years ago
  15. polpak
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    And be sure to group up any squares you find

    • 3 years ago
  16. kathyworld09
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    1,5,9,45 = factors for 45

    • 3 years ago
  17. polpak
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    Ignore 1 and itself.

    • 3 years ago
  18. kathyworld09
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    and for 5 , it would be just 5

    • 3 years ago
  19. polpak
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    45 = 9*5, and 9 is what?

    • 3 years ago
  20. kathyworld09
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    3*3

    • 3 years ago
  21. polpak
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    \(3^2\)

    • 3 years ago
  22. polpak
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    Ok, so lets rewrite the problem now with the squares.. \(\sqrt{5} - 7\sqrt{2^2 * 2^2 * 5} + 4\sqrt{3^2 * 5}\)

    • 3 years ago
  23. polpak
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    Now any square factor we can pull out of the square root by removing the square.

    • 3 years ago
  24. polpak
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    \(7*\sqrt{2^2 * 2^2 * 5} = 7 * \sqrt{2^2} * \sqrt{2^2} * \sqrt{5} = ?\)

    • 3 years ago
  25. kathyworld09
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    i multiply the sqrts ?

    • 3 years ago
  26. polpak
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    No, take the square root of the squared term. What is \(\sqrt{2^2}\)

    • 3 years ago
  27. polpak
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    If a >= 0, then \(\sqrt{a^2} = a\)

    • 3 years ago
  28. polpak
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    Are you still with me?

    • 3 years ago
  29. kathyworld09
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    yes

    • 3 years ago
  30. polpak
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    Ok, great. So what's \(\sqrt{2^2}\)

    • 3 years ago
  31. kathyworld09
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    sqrt. 4

    • 3 years ago
  32. polpak
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    Yes, but what's sqrt of 4? It's 2 right? You don't need to evaluate \(2^2\) because when you take the square root you just take off the square and the square root. They undo eachother. \(\sqrt{2^2} = 2. \) \(\sqrt{3^3} = 3.\) etc.

    • 3 years ago
  33. polpak
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    Err \(\sqrt{3^2} = 3\)

    • 3 years ago
  34. kathyworld09
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    so it would just be 2 and 3 ?

    • 3 years ago
  35. polpak
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    Not quite. So we have \(7*\sqrt{80} = 7*\sqrt{2^2 *2^2 * 5} = 7*\sqrt{2^2}*\sqrt{2^2} * \sqrt{5} =?\)

    • 3 years ago
  36. kathyworld09
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    7 * 2 sqrt. 5 ?

    • 3 years ago
  37. kathyworld09
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    correction : 7 * 2 * 2 * sqrt. 5 ?

    • 3 years ago
  38. polpak
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    Right. So now simplify that

    • 3 years ago
  39. polpak
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    7*2*2 = ?

    • 3 years ago
  40. kathyworld09
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    becuase their are 2 [sqrt.2^2]

    • 3 years ago
  41. polpak
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    No, we got rid of the square root already on those. \((7*2*2)\sqrt{5} \) \(= (14*2 ) \sqrt{5}\) \(= 28 \sqrt{5}\)

    • 3 years ago
  42. kathyworld09
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    can we still simplfy that ?

    • 3 years ago
  43. polpak
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    No, but we have \(\sqrt{5} - 28\sqrt{5} + 4\sqrt{45}\) So if we can get the \(4\sqrt{45}\) to be a multiple of \(\sqrt{5}\) we can combine them all.

    • 3 years ago
  44. polpak
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    So lets look at \(\sqrt{45}\). We said that that was 45 = 9*5 = \(3^2*5\), so what would \(4*\sqrt{45}\) simplify to?

    • 3 years ago
  45. kathyworld09
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    2 sqrt. 5 ?

    • 3 years ago
  46. kathyworld09
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    4 sqrt . 5 *3^2

    • 3 years ago
  47. polpak
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    So what is the next step now that you have that?

    • 3 years ago
  48. polpak
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    Just like the last one, we pull out each squared factor.

    • 3 years ago
  49. kathyworld09
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    4*3 sqrt.5 ?

    • 3 years ago
  50. polpak
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    Yes. Which is?

    • 3 years ago
  51. polpak
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    What's 4*3?

    • 3 years ago
  52. kathyworld09
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    12 sqrt. 5

    • 3 years ago
  53. polpak
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    Right. So now we have \(1\sqrt{5} - 28\sqrt{5} + 12\sqrt{5}\)

    • 3 years ago
  54. kathyworld09
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    -15 sqrt. 5 ?

    • 3 years ago
  55. polpak
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    Yes.

    • 3 years ago
  56. kathyworld09
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    and for the 1st problem , i factor out which is a ?

    • 3 years ago
  57. polpak
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    Do the same thing. Factor everything under the square root. Pull out any square factors by removing the square, leave the rest under the square root. Then combine.

    • 3 years ago
  58. polpak
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    Go ahead and do it yourself, step by step and I'll point out when/if you get off track.

    • 3 years ago
  59. polpak
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    Just post each step as you go.

    • 3 years ago
  60. polpak
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    What's the first step?

    • 3 years ago
  61. kathyworld09
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    [4^3 sqrt. a] + [^3 sqrt. 64a] : what do i do with the powers before the sqrt ?

    • 3 years ago
  62. kathyworld09
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    are you still here ?

    • 3 years ago
  63. polpak
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    I don't understand.. Do you have cube roots? \(4\sqrt[3]{a} + \sqrt[3]{64a}\) If so then instead of squares, you're looking for cubes. \(\sqrt[3]{2^3} = 2 \) is an example you will probably find useful. If you are still working on this, give it a try and write out your steps and where you're stuck and I'll check back again later.

    • 3 years ago
  64. kathyworld09
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    yes i still need help please

    • 3 years ago
  65. kathyworld09
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    polpak: i think it'll result to 16\[^{3} \sqrt{a}\]

    • 3 years ago
  66. polpak
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    What was the first step you did?

    • 3 years ago
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