How do you simplify #sqrt(75a^12b^3c^5)#?

1 Answer
Feb 4, 2015

We'll need a few properties.

  1. First of all, let's recall that #\sqrt{a\cdot b}=\sqrt{a} \sqrt{b}#. This of course applies also to products of more than two factors.
  2. I hope you will be ok with the fact that the square root of a number is that number to the power of #1/2#. If not, tell me in the comments and I will explain this exercise in another way.
  3. The third properties is that #a^{b+c}=a^b\cdot a^c#.

If this things are ok, for the first point we can separate the roots:
#\sqrt{75a^12b^3c^5}=\sqrt{75}\sqrt{a^12}\sqrt{b^3}\sqrt{c^5}#

Factoring #75# in prime numbers, we have #\sqrt{75}=\sqrt{3\cdot 5^2}=\sqrt{3}\sqrt{5^2}=5\sqrt{3}#.

For #\sqrt{a^12}#, we have that #\sqrt{a^12}=a^{12/2}=a^6#.

For #\sqrt{b^3}#, we have that #\sqrt{b^3}=b^{3/2}=b^{1+{1}/{2}}=b\sqrt{b}#

For #\sqrt{c^5}#, we have that #\sqrt{c^5}=c^{5/2}=c^{2+{1}/{2}}=c^2\sqrt{c}#.

Putting all the pieces together, we have that
#\sqrt{75a^12b^3c^5} = 5a^6bc^2\sqrt{3bc}#