For which value(s) of \( k \) will the equation \( x^{2}-5 x-k=0 \) have: \( \begin{array}{lll}\text { (a) equal roots } & \text { (b) real roots } & \text { (c) non-real roots }\end{array} \)

To determine the values of \( k \) for different types of roots in the quadratic equation \( x^{2} - 5x - k = 0 \), we can use the discriminant \( D \), which is given by \( D = b^2 - 4ac \). In this case, \( a = 1 \), \( b = -5 \), and \( c = -k \). 1. **Equal roots**: For the equation to have equal roots, the discriminant must be zero: \[ D = (-5)^2 - 4 \cdot 1 \cdot (-k) = 25 + 4k = 0 \] Solving for \( k \), we find: \[ 4k = -25 \quad \Rightarrow \quad k = -\frac{25}{4} \] 2. **Real roots**: For the equation to have real roots, the discriminant must be greater than or equal to zero: \[ 25 + 4k \geq 0 \] Solving for \( k \), we get: \[ 4k \geq -25 \quad \Rightarrow \quad k \geq -\frac{25}{4} \] 3. **Non-real roots**: For the equation to have non-real roots, the discriminant must be less than zero: \[ 25 + 4k < 0 \] Solving for \( k \), we find: \[ 4k < -25 \quad \Rightarrow \quad k < -\frac{25}{4} \] In summary: - Equal roots occur at \( k = -\frac{25}{4} \). - Real roots occur for \( k \geq -\frac{25}{4} \). - Non-real roots occur for \( k < -\frac{25}{4} \).