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Loujoelou

Simplify the problem below

  • one year ago
  • one year ago

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  1. Loujoelou
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    \[\sqrt{128}+2\sqrt{112}-2\sqrt{98}-6\sqrt{28}\] When I did this problem I got an answer not there so yeah the choices are: \[-6\sqrt{2}-4\sqrt{7}\] \[-12-4\sqrt{14}\] \[-4\sqrt{114}\] \[-30\]

    • one year ago
  2. Loujoelou
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    lol I've tried to do it 3 times but get the same thing.

    • one year ago
  3. tcarroll010
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    For a start, factor out the numbers within the radicals. Hint: 128 is a certain power of 2, that is, 2^(some number) = 128. Some of these factors, if the # of times multiplied is even, will come "out" of the radical. So, start with 128 and factor.

    • one year ago
  4. Loujoelou
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    When I simplify all of them to combine like terms I have- 6sqrt"2" + 4sqrt"7" - 7sqrt"2" - 12sqrt"7"

    • one year ago
  5. tcarroll010
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    That's where your problem is, but we can correct that. Do it in steps. How many times does 2 go into 128?

    • one year ago
  6. Loujoelou
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    128=2^7

    • one year ago
  7. tcarroll010
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    Yes! Good!, So, 2x2x2x2x2x2x2 = 128 and sqrt(2x2x2x2x2x2x2) = sqrt(128). You have an odd # of factors of "2", so one 2 will stay in the sqrt and 6 will come out. Observe: sqrt[(2x2)x(2x2)x(2x2)x2] = sqrt(128). Now the 2's come out as a "pair". There are 3 pairs. So, 2x2x2x[sqrt(2)] = sqrt(128)

    • one year ago
  8. tcarroll010
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    So, 8[sqrt(2)] = sqrt(128) The key to this problem is taking out factors in pairs and counting only ONE per pair. Making sense?

    • one year ago
  9. Loujoelou
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    yep

    • one year ago
  10. Loujoelou
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    wait why would 8sqrt"2"= sqrt128 if 128 only has 7 2's?

    • one year ago
  11. tcarroll010
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    The # of pairs is 3 (with a 2 left over), so 6 2's come out as 3 pairs. But at this point, you don't add 2+2+2, you multiply 2x2x2.

    • one year ago
  12. Loujoelou
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    Oh LOL i see my mistake now -_- careless mistake i didn't even realize lol thx :D

    • one year ago
  13. tcarroll010
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    They really are all factors. Another way to look at this is: Sqrt(128) = sqrt(2x2x2x2x2x2x2) = sqrt(2)sqrt(2)sqrt(2)sqrt(2)sqrt(2)sqrt(2)sqrt(2) = [sqrt(2)sqrt(2)][sqrt(2)sqrt(2)][sqrt(2)sqrt(2)]sqrt(2) = 2x2x2x sqrt(2)

    • one year ago
  14. tcarroll010
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    Sounds like you understand this pretty well now. All good now?

    • one year ago
  15. Loujoelou
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    so now it's 8sqrt"2" + 4sqrt"7" - 7sqrt"2" - 2sqrt"7" right?

    • one year ago
  16. tcarroll010
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    You have 4 terms in your last post. We now know the first one. Looking at just the second, you pulled out 2 pairs of 2's, plus you had a 2 in front already, so that second term should be 2x2x2x sqrt(7), not 4 x sqrt(7)

    • one year ago
  17. tcarroll010
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    Similar with the third term, there was a factor of 2 outside already, so that third term has to be multiplied by 2. Similar to the fourth term: that one has to be multiplied by the "6" that is outside already.

    • one year ago
  18. Loujoelou
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    oh yeah so the 2nd-4th term would be 8sqrt"7"+14sqrt"2"+12sqrt"7" right?

    • one year ago
  19. tcarroll010
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    You are extremely close to the answer. You have to take into account the signs. #3 and #4 terms have negative factors. Then, once you pute the right signs in, you combine the first and 3rd terms by addition (subtraction) because they are of the same sqrt(2). Similar for terms 2 and 4.

    • one year ago
  20. Loujoelou
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    okay I see so it'll be A, -6sqrt"2" - 4sqrt"7"

    • one year ago
  21. tcarroll010
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    \[a \times \sqrt{x}+b \times \sqrt{x} = (a +b)\times \sqrt{x}\]

    • one year ago
  22. tcarroll010
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    Very Good! Nice work! I went through all the steps, because now you will be able to do these on your own better. Isn't that a good sense of accomplishment?

    • one year ago
  23. Loujoelou
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    lol yeah I guess & I'm happy I got it right now and not drasitcally wrong like I had earlier without realizing

    • one year ago
  24. tcarroll010
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    You'll be fine now. Sometimes, getting through one stumbling block, gets us through a whole bunch since we conceptualize better now. You'll do fine.

    • one year ago
  25. Loujoelou
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    Would you mind helping me with another but this time it's rationalizing the denominator?

    • one year ago
  26. tcarroll010
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    Okay.

    • one year ago
  27. Loujoelou
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    It's not letting me go to my new question.

    • one year ago
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