Question 181557: I really need help on square roots, i dont get it. Well i have two problems that i was unable to solve..PLEASE HELP!!!
First: sqrt(162)
Well i tried dividing it by 6, and then did 6 sqrt(27)but the only problem is that its not right. And I know that because sqrt (162) and 6 sqrt (27) decimals are both different.
Second: 4 sqrt (48) + 5 sqrt (108)
For this one, i understand that the numbers under the square root have to be the same, and then add the outside, but i dont know how to get them the same inorder to add the outside.
Please Help, i would really appreciate it. AND its due on FRIDAY and today is wednesday! PLEASE HELP ME!! Thank You sooo much!
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443)   (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Actually, you shouldn't expect your method to work because:
(27)} = sqrt{6}sqrt{27} \neq 6 sqrt{27})
What you need to do first is find the prime factorization of the radicand. 162 is small enough that it won't be too tedious to do this with the brute force method.
 . Integer result, so one factor of 2.
81 is odd, so there are no more factors of 2, but 8 + 1 = 9, divisible by 3, so 81 is divisible by 3.
 . Integer result, so one factor of 3.
2 + 7 = 9, divisible by 3, so 27 is divisible by 3.
 . Integer result, so two factors of 3.
9 is divisible by 3.
 . Integer result, so three factors of 3, and the result is 3 so four factors of three.
Therefore, the prime factorization of  .
Now let's go back and look at your problem:
For every pair of prime factors, you can take one of the factors out of the radicand. That's because for a pair of prime factors, one of the factors is the square root of the product of the pair.
You have two pairs of 3s, so eliminate all of the 3s from under the radicand and take two 3s outside of the radical sign.
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Same process as above. Find the prime factorization of 48. Then find the prime factorization of 108. You should end up with both terms having a radicand of 3.
John
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