For what value of k are the roots of the equadratic equation (k+1)x2-2(k-1)x+1=0 real and equal? k के किस मान के लिए, द्विघात समीकरण (k+1)x2-2(k-1)x+1=0 के मूल वास्तविक और समान होंगें?

For real and equal roots, discriminant D = 0. Given equation: (k+1)x² - 2(k-1)x + 1 = 0. D = 4(k-1)² - 4(k+1)(1) = 0. Simplifying: 4[(k-1)² - (k+1)] = 0 → (k² - 2k + 1) - k - 1 = 0 → k² - 3k = 0 → k(k-3) = 0. Thus k = 0 or k = 3. We must also ensure k+1 ≠ 0 (coefficient of x²), so k ≠ -1. Both k=0 and k=3 satisfy this. Option C correctly states k=0 or k=3. Option D incorrectly suggests k=-3 as an alternative, which doesn't satisfy the equation.