## Loujoelou 2 years ago Simplify the problem below

1. Loujoelou

$\sqrt{128}+2\sqrt{112}-2\sqrt{98}-6\sqrt{28}$ 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

lol I've tried to do it 3 times but get the same thing.

3. tcarroll010

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

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

5. tcarroll010

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

128=2^7

7. tcarroll010

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

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

yep

10. Loujoelou

wait why would 8sqrt"2"= sqrt128 if 128 only has 7 2's?

11. tcarroll010

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

Oh LOL i see my mistake now -_- careless mistake i didn't even realize lol thx :D

13. tcarroll010

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

Sounds like you understand this pretty well now. All good now?

15. Loujoelou

so now it's 8sqrt"2" + 4sqrt"7" - 7sqrt"2" - 2sqrt"7" right?

16. tcarroll010

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

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

oh yeah so the 2nd-4th term would be 8sqrt"7"+14sqrt"2"+12sqrt"7" right?

19. tcarroll010

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

okay I see so it'll be A, -6sqrt"2" - 4sqrt"7"

21. tcarroll010

$a \times \sqrt{x}+b \times \sqrt{x} = (a +b)\times \sqrt{x}$

22. tcarroll010

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

lol yeah I guess & I'm happy I got it right now and not drasitcally wrong like I had earlier without realizing

24. tcarroll010

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

Would you mind helping me with another but this time it's rationalizing the denominator?

26. tcarroll010

Okay.

27. Loujoelou

It's not letting me go to my new question.