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Loujoelou

  • 2 years ago

Simplify the problem below

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  1. Loujoelou
    • 2 years ago
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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: \[-6\sqrt{2}-4\sqrt{7}\] \[-12-4\sqrt{14}\] \[-4\sqrt{114}\] \[-30\]

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

  3. tcarroll010
    • 2 years ago
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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.

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

  5. tcarroll010
    • 2 years ago
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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?

  6. Loujoelou
    • 2 years ago
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    128=2^7

  7. tcarroll010
    • 2 years ago
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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)

  8. tcarroll010
    • 2 years ago
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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?

  9. Loujoelou
    • 2 years ago
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    yep

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

  11. tcarroll010
    • 2 years ago
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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.

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

  13. tcarroll010
    • 2 years ago
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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)

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

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

  16. tcarroll010
    • 2 years ago
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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)

  17. tcarroll010
    • 2 years ago
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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.

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

  19. tcarroll010
    • 2 years ago
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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.

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

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

  22. tcarroll010
    • 2 years ago
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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?

  23. Loujoelou
    • 2 years ago
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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

  24. tcarroll010
    • 2 years ago
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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.

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

  26. tcarroll010
    • 2 years ago
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    Okay.

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

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