SOLUTION: simplify root symbol 39a^3b^5/root symbol 13a2b I am having difficulty with this can you help please????

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Question 623092: simplify
root symbol 39a^3b^5/root symbol 13a2b
I am having difficulty with this can you help please????

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I assume by "root symbol" you mean square root. (In the future, please specify which kind of root.) I also assume that "13a2b" means 13a%5E2%2Ab

There are several ways to deal with a fraction of square roots. My prefernce is to do the following:
  1. Use the root%28a%2C+p%29%2Froot%28a%2C+q%29+=+root%28a%2C+p%2Fq+%29 property to join the two square roots into one.
  2. Reduce the fraction. (Note: If you're clever you will keep in mind step 3 and maybe not reduce the fraction as fully as possible. For example, if you had sqrt%282%2F4%29 then you would not reduce the fraction because step 3 wants perfect square denominators. 4 is already a perfect square. If we reduce 2/4 to 1/2 then we would no longer have a perfect square denominator. And we would end up changing 1/2 back to 2/4 in step 3.)
  3. If there still is a denominator, then make sure it is a perfect square.
  4. If there still is a denominator, then use the sqrt%28p%29%2Fsqrt%28q%29+=+sqrt%28p%2Fq%29 property again, this time in the other direction -- to split the numerator from the denominator.
  5. Simplify any remaining square roots and fractions.
Let's see this in action:
sqrt%2839a%5E3b%5E5%29%2Fsqrt%2813a%5E2%2Ab%29
1. Join the square roots:
sqrt%28%2839a%5E3b%5E5%29%2F%2813a%5E2%2Ab%29%29
2. Reduce the fraction (maybe not fully).
The entire denominator cancels out leaving:
sqrt%28%283ab%5E4%29%2F1%29
or just
sqrt%283ab%5E4%29
3. If there's a still denominator, ...
There is no denominator.
4. If there's a still denominator, ...
There is no denominator.
5. Simplify any remaining square roots and fractions.
To simplify a square root, look for perfect square factors of the radicand (the expression inside). There are not prefect square factors in the 3 or the a. But there are perfect square factors in b%5E4. When I factor out the perfect squares I like to put those factors in front:
sqrt%28b%5E2%2Ab%5E2%2A3a%29
Then you use another property of radicals, root%28a%2C+p%2Aq%29+=+root%28a%2C+p%29%2Aroot%28a%2C+q%29, split the square root so that eah perfect square factor is in its own square root:
sqrt%28b%5E2%29%2Asqrt%28b%5E2%29%2Asqrt%283a%29
The square roots of the perfect squares simplify:
b%2Ab%2Asqrt%283a%29
which simplifies to:
b%5E2%2Asqrt%283a%29