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 which the quadratic equation \( x^{2}-5x-k=0 \) has equal roots, real roots, and non-real roots, we will use the discriminant of the quadratic equation. The discriminant \( D \) is given by the formula: \[ D = b^2 - 4ac \] For our equation \( x^{2}-5x-k=0 \), we have: - \( a = 1 \) - \( b = -5 \) - \( c = -k \) Thus, the discriminant becomes: \[ D = (-5)^2 - 4(1)(-k) = 25 + 4k \] Now, we will analyze the discriminant for each case: ### (a) Equal Roots For the equation to have equal roots, the discriminant must be equal to zero: \[ D = 0 \implies 25 + 4k = 0 \] Solving for \( k \): \[ 4k = -25 \implies k = -\frac{25}{4} \] ### (b) Real Roots For the equation to have real roots, the discriminant must be greater than or equal to zero: \[ D \geq 0 \implies 25 + 4k \geq 0 \] Solving for \( k \): \[ 4k \geq -25 \implies k \geq -\frac{25}{4} \] ### (c) Non-Real Roots For the equation to have non-real roots, the discriminant must be less than zero: \[ D < 0 \implies 25 + 4k < 0 \] Solving for \( k \): \[ 4k < -25 \implies k < -\frac{25}{4} \] ### Summary of Results - (a) Equal roots: \( k = -\frac{25}{4} \) - (b) Real roots: \( k \geq -\frac{25}{4} \) - (c) Non-real roots: \( k < -\frac{25}{4} \)