assembly - ASM8086: mul, imul, carry flag and overflow flag -

i understood logic of carry flag , overflow flag. but, when read program (wrote in masm 8086) , got perplexed bit.

the program intent tell if quadratic equation has 2 distincts solutions, 2 equals solutions or no solutions @ all.

.model small .stack .data aa dw 2 bb dw 4 cc dw 2 sol_msg    db "there exist 2 real solutions", cr, nl no_sol_msg db "no real solutions!            ", cr, nl sol_coinc  db "the 2 solutions coincide!   ", cr, nl .code .startup mov ax, bb imul bb jc overflow ; decided work @ 16-bit numbers push ax mov ax, aa imul cc jc overflow mov bx, 4 imul bx jc overflow pop bx sub bx, ax jo overflow js mess2 jz mess3 lea si, sol_msg jmp next mess2: lea si, no_sol_msg jmp next mess3: lea si, sol_coinc next: mov bx, lung_msg mov ah, 2 loop1: mov dl, [si] int 21h inc si dec bx jnz loop1 jmp end1 overflow: nop end1:  .exit end 

now, doubt is: why first 3 checks has been tested carry flag , last 1 overflow flag?

since in last one, subtraction between 2 signed numbers (as think those), have check overflow flag see if there's overflow (i.e. if numbers goes beyond interval[-2^15,2^15-1]). first one, example, (bb)^2 imul.

so think them 16-bit signed numbers (so -2^15 <= bb <= 2^15-1 ) and, multiplication set cf/of bit on if @ least 1 sum (in multiplication algorithm) cf/of bit on.

but since treat signed numbers, shouldn't check overflow flag ?

also noticed that, since 2^15-1=32767, if set bb 190 (190^2=36100) cf=0; if bb equals 200 (200^2=40000) cf=1.

why that? explain me in detailed way, please?

p.s.: i'm using emu8086.

if have @ pseudo algorithm of imul, you'll see

if operandsize = 16   tmp_xp ← ax ∗ src (* signed multiplication; tmp_xp signed integer @ twice width of src *)   dx:ax ← tmp_xp[31:0];   sf ← tmp_xp[15];   if signextend(tmp_xp[15:0]) = tmp_xp     cf ← 0; of ← 0;     else cf ← 1; of ← 1;   fi; fi; 

cf , of either both set, or both cleared. so, can check either flag after imul