how to use distributive property with $\sqrt{x-2}$?

Question

with exponents it can be done $(x-2) ^ 2 = (x-2) * (x-2)$

and then apply distributive property

how to do it with $\sqrt{x-2}$?

Answer 1

Wolfram Alpha is unable to expand it in the way you're thinking, but it does give some possible series expansions at $x=0$, $x=\pi$ and $x=\infty$ using Puiseaux and Taylor series. My own research has led me to conclude that the distributive property does not hold for polynomials raised to fractional exponents, such as your example.

Answer 2

To apply the distributive property, we must be able to distribute (multiply) one of the factors, in the first case $(x-2)$, over the other set of parenthesis. This is possible in the first example because expanding any squared binomial yields 2 binomials each raised to the power of 1. It is only for this reason, because 1 is the multiplicative and exponential identity, that we are able to apply multiplication by distribution at this step, because we have eliminated the exponents:

$$ (x-2)(x-2)=x(x-2)-2(x-2)=x^2-4x+4 $$

Let us treat your second case identically to how you began your first:

$$ \sqrt{x-2}=(x-2)^\frac{1}{2}=(x-2)^\frac{1}{4}(x-2)^\frac{1}{4} $$

Here we cannot apply the distributive property because the exponents have not been eliminated. There is nothing to distribute. It is a special property, then, of a squared binomial, for us to be able to "bypass" the exponents by halving the power.