To solve for the expression \( x + x^{-1} \), we can rewrite it in a more manageable form. The expression can also be seen as:
\[ x + \frac{1}{x} \]
To find solutions, we can set up an equation. Typically, one might be interested in setting this equal to a specific value (let's say \( k \)), so we have:
\[ x + \frac{1}{x} = k \]
To solve for \( x \), we can multiply both sides of the equation by \( x \) (assuming \( x \neq 0 \)):
\[ x^2 + 1 = kx \]
Rearranging gives:
\[ x^2 - kx + 1 = 0 \]
This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -k \), and \( c = 1 \). To solve this quadratic, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting in our values of \( a \), \( b \), and \( c \):
\[ x = \frac{k \pm \sqrt{k^2 - 4}}{2} \]
The solutions depend on the value of \( k \). The expression \( x + x^{-1} \) has specific solutions for different values of \( k \). For example:
- If \( k = 2 \), then \( x + x^{-1} = 2 \) gives \( x = 1 \) or \( x = -1 \).
- If \( k < -2 \), there are no real solutions since the discriminant would be negative.
If you have a specific value of \( k \) in mind, please provide it so I can present a more tailored solution!
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