Two solutions for a given problem when simplifying double radicals?

Question

Given the following problem:

Simplify: $$\sqrt{8-2 \times \sqrt{15}}$$

This expression could be rewritten as:

$$\begin{align}\\ \sqrt{8-2 \times \sqrt{15}}&= \sqrt{5-2\times\sqrt{5\times3}+3}\\ &=\sqrt{(\sqrt{5} - \sqrt{3})^2}\end{align}$$

Now since this expression is in the form $\sqrt{a^2} = \lvert a \rvert$

It follows that both $\sqrt{5} - \sqrt{3}$ and $\sqrt{3} - \sqrt{5}$ satisfy the expression above because they are equal to $a$ and $-a$ respectively.

The author (Kunihiko Kodaira, Japanese Grade 10) selects $\sqrt{5}-\sqrt{3}$ as a correct answer.

Where is the flaw in my logic?

Answer 1

Hint:

$ \sqrt{8-2\sqrt{5}}>0$ by definition

so is is equal to $\sqrt{5}-\sqrt{3} >0$

Answer 2

$|\sqrt{5}-\sqrt{3}|=\sqrt{5}-\sqrt{3}$ only.

$|\sqrt{5}-\sqrt{3}| \neq \sqrt{3}-\sqrt{5}$

Answer 3

With real numbers, the square (positive) root of a square is the absolute value of the base of that square.

The fact that the "square goes with the root" we say it to the children.

Your logic is right, both are correct answers.

They probably saved the other answer to save on paper (or in bits)