Question

what is the simplfied form of sqrt(72n^13) and
sqrt(k^17)?

Answer 1

\[\sqrt{72n^{13}}\]\[\sqrt{k^{17}}\]
For the first one. You can split the 72 and n^13 up so it's like this:\[\sqrt{72n^{13}} = \sqrt{(36 \times 2)(n^{12} \times n)}\]
Then you know the sqrt of 36 is 6, and the sqrt of n^12 is n^6, so you take those out and you're left with \[6n^6 \sqrt{2n}\]
For the second, you do the same thing we did with the n in the first example:
\[\sqrt{k^{17}} = \sqrt{k^{16} \times k}\] and the sqrt of k^16 is obviously k^8, take that out and you're left with \[k^8 \sqrt{k}\]