2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.(i) 2x2 + kx + 3 = 0(ii) kx (x – 2) + 6 = 0

(i) 2x2 + kx + 3 = 0 Comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = k and c = 3 As we know, Discriminant = b2 – 4ac = (k)2 – 4(2) (3) = k2 – 24 For equal roots, we know, Discriminant = 0 k2 – 24 = 0 k2 = 24 k = ±√24 = ±2√6 (ii) kx(x – 2) + 6 = 0 or kx2 – 2kx + 6 = 0 Comparing the given equation with ax2 + bx + c = 0, we get a = k, b = – 2k and c = 6 We know, Discriminant = b2 – 4ac = ( – 2k)2 – 4 (k) (6) = 4k2 – 24k For equal roots, we know, b2 – 4ac = 0 4k2 – 24k = 0 4k (k – 6) = 0 Either 4k = 0 or k = 6 = 0 k = 0 or k = 6 However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘. Therefore, if this equation has two equal roots, k should be 6 only.