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

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

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We rewrite the equation as $$ x^2 - x + 1 - \frac{1}{k} = 0. $$ This is a quadratic in $$x$$ of the form $$ ax^2 + bx + c = 0, $$ with $$ a = 1,\quad b = -1,\quad c = 1 - \frac{1}{k}. $$ For the quadratic to have real roots, the discriminant must be nonnegative: $$ \Delta = b^2 - 4ac \ge 0. $$ Substituting the values, we get $$ \Delta = (-1)^2 - 4 \cdot 1 \left(1 - \frac{1}{k}\right) = 1 - 4 + \frac{4}{k} = \frac{4}{k} - 3. $$ Thus, the condition for real roots is $$ \frac{4}{k} - 3 \ge 0. $$ Solving the inequality, we first isolate $$\frac{4}{k}$$: $$ \frac{4}{k} \ge 3. $$ Since $$k \neq 0$$, we consider the sign of $$k$$. **Case 1: $$k > 0$$** Multiply both sides by $$k$$ (which is positive, so the inequality sign remains the same): $$ 4 \ge 3k \quad \Longrightarrow \quad k \le \frac{4}{3}. $$ Thus, for $$k > 0$$ the inequality becomes $$ 0 < k \le \frac{4}{3}. $$ **Case 2: $$k < 0$$** If $$k$$ is negative, then $$\frac{4}{k}$$ is negative. The inequality $$ \frac{4}{k} \ge 3 $$ cannot hold since $$3$$ is positive. Therefore, there are no solutions when $$k < 0$$. Thus, the equation has real roots if and only if $$ 0 < k \le \frac{4}{3}. $$ The equation $$ \frac{1}{k} = x^{2} - x + 1 $$ has real roots when $$ 0 < k \leq \frac{4}{3} $$.

Determine the value(s) of k for which the equation
where has real roots.

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Determine the value(s) of k for which the equation where has real roots.