Let p and q be the real roots of the equation x² - 4x + A = 0 and let r and s be the real roots of the equation x² - 12x + B = 0, where A and B are positive integers. If p, q, r, s are in arithmetic progression and p < q < r < s, then find the value of A*B.

First equation, x² - 4x + A = 0 Discriminant D = 16 - 4A Roots are (4 - (16 - 4A)^1/2)/2 and (4 + (16 - 4A)^1/2)/2, i.e. the roots are (2-(4-A)^1/2)and (2+(4-A)^1/2). Obviously, q = (2+(4-A)^1/2) and p = (2-(4-A)^1/2) because it is written that q>p. So, q-p = 2(4-A)^1/2 Second equation, x² - 12x + B =0 Discriminant D = 144 - 4B Roots are (12 - (144 - 4B)^1/2)/2 and (12 + (144 - 4B)^1/2)/2, i.e. the roots are (6-(36-B)^1/2)and (6+(36-B)^1/2). Obviously s = (6+(36-B)^1/2) and r = (6-(36-B)^1/2) because it is written that s>r. So, s-r = 2(36-B)^1/2 Since p, q, r and s are in AP, therefore q - p = s - r (In AP, common difference is the same) 2(4-A)^1/2 = 2(36-B)^1/2 Solving, we get 4-A = 36-B, i.e. B-A = 32 Now, possible combinations of (B,A) are (36,4), (35,3), (34,2), (33,1) because A<=4 as per the roots of first equation {(4-A)^1/2}, otherwise roots will be complex. Also, B<=36 as per the roots of the second equation. Hence, this option is correct.