Find the values of k for which (k + 1)x² - 2(k - 1)x + 1 = 0 has two equal roots…

Step 1. The discriminant. $D = 4(k - 1)^2 - 4(k + 1) = 4(k^2 - 2k + 1 - k - 1) = 4(k^2 - 3k)$.

Step 2. Set it to zero. $k^2 - 3k = 0$, so $k(k - 3) = 0$, and $k = 0$ or $k = 3$.

Step 3. The root for k = 3. The equation is $4x^2 - 4x + 1 = 0$, that is $(2x - 1)^2 = 0$, so $x = \dfrac{1}{2}$.