Show that the equation \( x^{2}+(k-2) x-k=0 \) has real roots for all real values of \( k \text {. } \)

To determine whether the equation \( x^{2}+(k-2) x-k=0 \) has real roots for all real values of \( k \), we need to analyze the discriminant of the quadratic equation. The general form of a quadratic equation is given by: \[ ax^2 + bx + c = 0 \] where \( a = 1 \), \( b = k - 2 \), and \( c = -k \) in our case. The discriminant \( D \) of a quadratic equation is calculated using the formula: \[ D = b^2 - 4ac \] Substituting the values of \( a \), \( b \), and \( c \): \[ D = (k - 2)^2 - 4(1)(-k) \] Now, let's simplify this expression step by step: 1. Calculate \( (k - 2)^2 \): \[ (k - 2)^2 = k^2 - 4k + 4 \] 2. Calculate \( -4(1)(-k) \): \[ -4(1)(-k) = 4k \] 3. Combine these results to find \( D \): \[ D = k^2 - 4k + 4 + 4k \] \[ D = k^2 + 4 \] Now, we observe that \( D = k^2 + 4 \). Since \( k^2 \) is always non-negative for all real values of \( k \) (i.e., \( k^2 \geq 0 \)), it follows that: \[ D \geq 4 \] This means that the discriminant \( D \) is always greater than or equal to 4, which is a positive number. Since the discriminant is positive, the quadratic equation \( x^{2}+(k-2) x-k=0 \) has two distinct real roots for all real values of \( k \). Thus, we have shown that the equation has real roots for all real values of \( k \).