Here's the question you clicked on:
Loujoelou
Simplify the problem below
\[\sqrt{128}+2\sqrt{112}-2\sqrt{98}-6\sqrt{28}\] When I did this problem I got an answer not there so yeah the choices are: \[-6\sqrt{2}-4\sqrt{7}\] \[-12-4\sqrt{14}\] \[-4\sqrt{114}\] \[-30\]
lol I've tried to do it 3 times but get the same thing.
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.
When I simplify all of them to combine like terms I have- 6sqrt"2" + 4sqrt"7" - 7sqrt"2" - 12sqrt"7"
That's where your problem is, but we can correct that. Do it in steps. How many times does 2 go into 128?
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)
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?
wait why would 8sqrt"2"= sqrt128 if 128 only has 7 2's?
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.
Oh LOL i see my mistake now -_- careless mistake i didn't even realize lol thx :D
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)
Sounds like you understand this pretty well now. All good now?
so now it's 8sqrt"2" + 4sqrt"7" - 7sqrt"2" - 2sqrt"7" right?
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)
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.
oh yeah so the 2nd-4th term would be 8sqrt"7"+14sqrt"2"+12sqrt"7" right?
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.
okay I see so it'll be A, -6sqrt"2" - 4sqrt"7"
\[a \times \sqrt{x}+b \times \sqrt{x} = (a +b)\times \sqrt{x}\]
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?
lol yeah I guess & I'm happy I got it right now and not drasitcally wrong like I had earlier without realizing
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.
Would you mind helping me with another but this time it's rationalizing the denominator?
It's not letting me go to my new question.