Given \[ x^{2}+2 x(k+1)+(2 k+1)=0 \] Show that the roots of the equation are real for all real values of \( k \).

To show that the roots of the quadratic equation \[ x^{2}+2 x(k+1)+(2 k+1)=0 \] are real for all real values of \( k \), we can use the discriminant method. The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \[ D = b^2 - 4ac. \] In our case, we identify \( a = 1 \), \( b = 2(k+1) \), and \( c = 2k + 1 \). Now, we compute the discriminant: \[ D = [2(k+1)]^2 - 4 \cdot 1 \cdot (2k + 1). \] Simplifying this, we have: \[ D = 4(k+1)^2 - 4(2k + 1) = 4[k^2 + 2k + 1 - (2k + 1)] = 4(k^2 + 2k + 1 - 2k - 1). \] This reduces to: \[ D = 4k^2. \] Notice that \( 4k^2 \) is always non-negative for all real \( k \) since the square of any real number is non-negative. Thus, \( D \geq 0 \). Therefore, the roots of the equation are real for all real values of \( k \). This concludes the demonstration.