Determine the value(s) of \( k \) for which the equation \( \frac{1}{k}=x^{2}-x+1 \) where \( k \neq 0 \) has real roots.

To determine the values of \( k \) for which the equation \( \frac{1}{k} = x^2 - x + 1 \) has real roots, we can first rearrange the equation into a standard quadratic form. By setting \( y = x^2 - x + 1 \), we can rewrite the equation as: \[ x^2 - x + (1 - \frac{1}{k}) = 0. \] This is a quadratic equation in the form \( ax^2 + bx + c = 0 \) where: - \( a = 1 \) - \( b = -1 \) - \( c = 1 - \frac{1}{k} \). For this quadratic to have real roots, the discriminant must be non-negative. The discriminant \( D \) is given by: \[ D = b^2 - 4ac = (-1)^2 - 4(1)(1 - \frac{1}{k}). \] Calculating this gives: \[ D = 1 - 4(1 - \frac{1}{k}) = 1 - 4 + \frac{4}{k} = \frac{4}{k} - 3. \] Setting the discriminant \( D \) to be greater than or equal to zero, we have: \[ \frac{4}{k} - 3 \geq 0. \] This simplifies to: \[ \frac{4}{k} \geq 3. \] To solve this inequality, we can multiply both sides by \( k \) (keeping in mind that \( k \neq 0 \) and the direction of the inequality will depend on the sign of \( k \)): 1. If \( k > 0 \): \[ 4 \geq 3k \implies k \leq \frac{4}{3}. \] 2. If \( k < 0 \): \[ 4 \leq 3k \implies k \geq \frac{4}{3} \text{ (but this is impossible for negative \( k \))}. \] Thus, combining our results, the only values of \( k \) for which the equation \( \frac{1}{k} = x^2 - x + 1 \) has real roots is when \( k \) is positive and \( k \leq \frac{4}{3} \). The final solution is: \[ 0 < k \leq \frac{4}{3}. \]