SOLUTION: Let a and B be solutions of the quadratic equation {{{x^2+bx+3=0}}}. Find all values of b such that {{{a^2+B^2=3}}}. Note: Capitalized B is not the same as lowercase b.

First, we have this identity

    a^2 + B^2 = (a + B)^2 - 2aB.    (1)


In this identity, we replace left side  a^2 + B^2 by 3;

next, we replace (a+B) by -b, according to Vieta's theorem;

and replace aB by 3, using the Vieta's theorem again.


We get then from (1)

    3 = (-b)^2 - 2*3,

which gives

    3 + 6 = b^2,

or 

    b^2 = 9.


ANSWER.  b^2 must be 9;  so "b" may have values 3 or -3.


As the last step of the solution, we chould check the problem's statement for b = 3 and b = -3.


It means 

    - (a) to find the solutions to equation x^2 + 3x + 3 = 0 and to check that 
          the sum of their squares is 3;

    - (b) to find the solutions to equation x^2 - 3x + 3 = 0 and to check that 
          the sum of their squares is 3.


Both steps are simple arithmetic, so I leave it for you.


After completing the check, we can state for sure that "b" may have two possible values 3 and -3.