Why doesn't √6 get divided by 6 at the end of the problem?

Question

So I'm working out the problem 3x^(2)-72, and the last two steps of the problem are

x=(+/- 12√6)/6

x=+/- 2√6 .   (+/- meaning the symbol for a positive or negative answer)

12 gets divided by 6, but √6 doesn't. I understand that √6 cannot be divided by 6, but why does the 6 disappear entirely? Why isn't it written as x=+/- 2 (√6/2)?

Answer 1

±12√6/6=(12√6)/6. The whole numerator is 12√6 and the denominator is just 6.

12√6=12×√6. When this is divided by 6, only one of these factors can be divided by 6. If 12 is divided by 6 we get 2 and the square root isn’t touched. If √6 is divided by 6 we get 1/√6, so we would be left with 12/√6. If it had been ±(12+√6)/6, then we would divide 6 into both components of the numerator.

Instead of √6 let’s suppose it was ±12x/6, the answer would be ±2x. So if x=√6, the answer would be ±2√6.

If your original question had been 3x²-72=0, then x²-24=0=(x-√24)(x+√24). 24=4×6, so √24=√4×√6=2√6 and x=±2√6.