To solve for [tex]\(\sqrt{x} + \frac{1}{\sqrt{x}}\)[/tex] given that [tex]\(x + \frac{1}{x} = 2\)[/tex], we can proceed as follows:
1. Start with the given equation:
[tex]\[
x + \frac{1}{x} = 2
\][/tex]
2. Let [tex]\(\sqrt{x} = y\)[/tex]. Therefore, [tex]\(x = y^2\)[/tex].
3. Substitute [tex]\(x = y^2\)[/tex] into the given equation:
[tex]\[
y^2 + \frac{1}{y^2} = 2
\][/tex]
4. Multiply both sides by [tex]\(y^2\)[/tex] to clear the fraction:
[tex]\[
y^4 + 1 = 2y^2
\][/tex]
5. Rearrange the equation to form a standard quadratic equation:
[tex]\[
y^4 - 2y^2 + 1 = 0
\][/tex]
6. Let [tex]\(z = y^2\)[/tex]. Substituting [tex]\(z\)[/tex] into the equation gives:
[tex]\[
z^2 - 2z + 1 = 0
\][/tex]
7. This is a perfect square quadratic and can be factored as:
[tex]\[
(z - 1)^2 = 0
\][/tex]
8. Solving for [tex]\(z\)[/tex], we get:
[tex]\[
z = 1
\][/tex]
9. Recall that [tex]\(z = y^2\)[/tex], so:
[tex]\[
y^2 = 1
\][/tex]
10. Solving for [tex]\(y\)[/tex], we obtain:
[tex]\[
y = \pm1
\][/tex]
11. Therefore, [tex]\(\sqrt{x} = 1\)[/tex] or [tex]\(\sqrt{x} = -1\)[/tex].
12. Now compute [tex]\(\sqrt{x} + \frac{1}{\sqrt{x}}\)[/tex]:
- If [tex]\(\sqrt{x} = 1\)[/tex], then [tex]\(\sqrt{x} + \frac{1}{\sqrt{x}} = 1 + \frac{1}{1} = 1 + 1 = 2\)[/tex].
- If [tex]\(\sqrt{x} = -1\)[/tex], then [tex]\(\sqrt{x} + \frac{1}{\sqrt{x}} = -1 + \frac{1}{-1} = -1 - 1 = -2\)[/tex]; however, this does not conform to the rule that the expression involves real numbers and positive and reciprocal nature.
Thus, the valid and conformational value of [tex]\(\sqrt{x} + \frac{1}{\sqrt{x}}\)[/tex] is [tex]\(2\)[/tex].
Therefore, the correct answer is [tex]\( \boxed{2} \)[/tex].
