Maths F4...13

19(a) (x + 3k)^2 + 2(3k + 2x) = 0x^2 + (6k + 4)x + (9k^2 + 6k) = 0Discriminant = (6k + 4)^2 - 4(9k^2 + 6k)= 36k^2 + 48k + 16 - 36k^2 - 24k= 24k + 16(b) 24k + 16 > 0 for all positive number kSo, the equation has real roots20 when k = 1The equation becomes 5x + 3 = 0which has one root only Otherwise, Discriminant= (2k + 3)^2 - 12(k - 1)= 4k^2 + 12k + 9 - 12k + 12= 4k^2 + 21> 0 And so has two real roots