If our two roots are [tex]r[/tex] and [tex]s[/tex] we can write
[tex]0 = (x-r)(x-s) = x^2 + bx + 6[/tex]
[tex]x^2 -(r+s)x+rs = x^2 + bx + 6[/tex]
Matching coefficients,
[tex]-(r+s) = b[/tex]
[tex]rs = 6[/tex]
"Positve integral" is a fancy way of saying natural number, 1, 2, 3, ...
So these are the factors pairs of six:
[tex]6=1 \cdot 6 = 2 \cdot 3[/tex]
The two possible sums are 1+6=7 and 2+3=5. b is the negative sum.
Answer: b=-7 or b=-5
