To solve for the positive value of [tex]\( \left(x + \frac{1}{x}\right) \)[/tex] given that [tex]\( x^2 + \frac{1}{x^2} = 23 \)[/tex], we can follow these steps:
1. Introduce a new variable:
Let [tex]\( y = x + \frac{1}{x} \)[/tex]. We aim to express [tex]\( x^2 + \frac{1}{x^2} \)[/tex] in terms of [tex]\( y \)[/tex].
2. Square the new variable:
[tex]\[
y^2 = \left(x + \frac{1}{x}\right)^2
\][/tex]
Using the binomial theorem to expand this expression, we get:
[tex]\[
y^2 = x^2 + 2 \cdot x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2
\][/tex]
Simplifying further:
[tex]\[
y^2 = x^2 + 2 + \frac{1}{x^2}
\][/tex]
3. Relate the given equation:
We are given that [tex]\( x^2 + \frac{1}{x^2} = 23 \)[/tex]. Substituting this into the equation:
[tex]\[
y^2 = 23 + 2
\][/tex]
Therefore:
[tex]\[
y^2 = 25
\][/tex]
4. Solve for [tex]\( y \)[/tex]:
[tex]\[
y^2 = 25 \implies y = \pm 5
\][/tex]
Since [tex]\( y = x + \frac{1}{x} \)[/tex] is stated to be positive, we select the positive value:
[tex]\[
y = 5
\][/tex]
Thus, the positive value of [tex]\( \left( x + \frac{1}{x} \right) \)[/tex] is [tex]\( \boxed{5} \)[/tex].
