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Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{128}+2\sqrt{112}2\sqrt{98}6\sqrt{28}\] When I did this problem I got an answer not there so yeah the choices are: \[6\sqrt{2}4\sqrt{7}\] \[124\sqrt{14}\] \[4\sqrt{114}\] \[30\]

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0lol I've tried to do it 3 times but get the same thing.

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1For 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.

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0When I simplify all of them to combine like terms I have 6sqrt"2" + 4sqrt"7"  7sqrt"2"  12sqrt"7"

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1That's where your problem is, but we can correct that. Do it in steps. How many times does 2 go into 128?

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1Yes! 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)

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1So, 8[sqrt(2)] = sqrt(128) The key to this problem is taking out factors in pairs and counting only ONE per pair. Making sense?

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0wait why would 8sqrt"2"= sqrt128 if 128 only has 7 2's?

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1The # 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.

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0Oh LOL i see my mistake now _ careless mistake i didn't even realize lol thx :D

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1They 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)

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1Sounds like you understand this pretty well now. All good now?

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0so now it's 8sqrt"2" + 4sqrt"7"  7sqrt"2"  2sqrt"7" right?

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1You 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)

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1Similar 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.

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0oh yeah so the 2nd4th term would be 8sqrt"7"+14sqrt"2"+12sqrt"7" right?

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1You 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.

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0okay I see so it'll be A, 6sqrt"2"  4sqrt"7"

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1\[a \times \sqrt{x}+b \times \sqrt{x} = (a +b)\times \sqrt{x}\]

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1Very 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?

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0lol yeah I guess & I'm happy I got it right now and not drasitcally wrong like I had earlier without realizing

tcarroll010
 2 years ago
Best ResponseYou've already chosen the best response.1You'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.

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0Would you mind helping me with another but this time it's rationalizing the denominator?

Loujoelou
 2 years ago
Best ResponseYou've already chosen the best response.0It's not letting me go to my new question.
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