SOLUTION: If sinx-cosx= 1 /√3 , and sinx and cosx are the two roots to the equation 2x^2+px+q=0, the value of p^2- 8q would be

If sin(x) and cos(x) are the two roots of the equation 2x^2+px+q=0, 

then due to Vieta's theorem

    sin(x) + cos(x) = -,     (1)

    sin(x)*cos(x) = .        (2)


Squaring equation (1) (both sides), it implies

    sin^2(x) + 2sin(x)*cos(x) + cos^2(x) = 
or
   2sin(x)*cos(x) =  - ,  

   sin(x)*cos(x) =  - 

     =  - 

    p^2 = 4q + 4

    p^2 - 8q = (4q +4) - 8q = -4q + 4.      (3)


Next, from the other given equality,

    sin(x) - cos(x) = .


Square both sides 

    sin^2(x) - 2sin(x)*cos(x) + cos^2(x) = 

    -2sin(x)*cos(x) =  - 1 = -

     sin(x)*cos(x) = .    (4)


Now from (1), (2), (3) and (4)

    p^2 - 8q = -4q + 4 = -4*(2sin(x)*cos(x)) + 4 =  + 4 = 4 -   =  = .


ANSWER.  Under given conditions,  p^2 - 8q = .