10. Find the value of \( k \) for which the equation \( (k-2) x^{2}+x(2 k+1)+k=0 \) has equal roots.

To determine the value of \( k \) for which the quadratic equation \[ (k-2)x^2 + (2k+1)x + k = 0 \] has equal roots, we need to use the condition for equal roots in a quadratic equation. A quadratic equation \( ax^2 + bx + c = 0 \) has equal roots if its discriminant \( D \) is zero, where the discriminant is given by \[ D = b^2 - 4ac. \] In our case, we identify the coefficients as follows: - \( a = k - 2 \) - \( b = 2k + 1 \) - \( c = k \) Now, we can compute the discriminant: \[ D = (2k + 1)^2 - 4(k - 2)(k). \] Expanding \( D \): 1. Calculate \( (2k + 1)^2 \): \[ (2k + 1)^2 = 4k^2 + 4k + 1. \] 2. Calculate \( 4(k - 2)(k) \): \[ 4(k - 2)(k) = 4(k^2 - 2k) = 4k^2 - 8k. \] Now, substituting these into the discriminant: \[ D = 4k^2 + 4k + 1 - (4k^2 - 8k). \] Simplifying this expression: \[ D = 4k^2 + 4k + 1 - 4k^2 + 8k = 12k + 1. \] To find the value of \( k \) for which the roots are equal, we set the discriminant equal to zero: \[ 12k + 1 = 0. \] Solving for \( k \): \[ 12k = -1 \quad \Rightarrow \quad k = -\frac{1}{12}. \] Thus, the value of \( k \) for which the equation has equal roots is \[ \boxed{-\frac{1}{12}}. \]