Given \[ 2 x^{2}+(k+1) x=-k+1 \] Show that the roots are real and rational for all \( k \in \mathbb{Z} \).

To determine if the roots of the quadratic equation \[ 2x^2 + (k + 1)x + (k - 1) = 0 \] are real and rational for all \( k \in \mathbb{Z} \), we can use the discriminant. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where the discriminant \( D \) is given by: \[ D = b^2 - 4ac \] In our case, \( a = 2 \), \( b = k + 1 \), and \( c = k - 1 \). Thus, the discriminant becomes: \[ D = (k + 1)^2 - 4(2)(k - 1) \] Now, let's simplify the discriminant: \[ D = (k + 1)^2 - 8(k - 1) \] \[ D = (k^2 + 2k + 1) - (8k - 8) \] \[ D = k^2 + 2k + 1 - 8k + 8 \] \[ D = k^2 - 6k + 9 \] \[ D = (k - 3)^2 \] The discriminant \( (k - 3)^2 \) is always non-negative because the square of any real number is non-negative. This means that the quadratic has real roots for any integer \( k \). Next, we need to check if the roots are rational. A quadratic equation has rational roots if its discriminant is a perfect square. In this case, \( (k - 3)^2 \) is indeed a perfect square for any integer \( k \). Therefore, since both conditions for the roots (real and rational) are satisfied, we can conclude that the roots of the equation \( 2x^2 + (k + 1)x + (k - 1) = 0 \) are real and rational for all \( k \in \mathbb{Z} \). This is a great example of how analyzing the structure of a mathematical statement can lead to profound and simple results!