Question

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

Answer 1

sinx and cosx are the two roots to the equation 2x^2+px+q=0, then
2(sinx)^2 + p sinx + q = 0 , and
2(cosx)^2 +p cosx + q = 0
subtract:
2sin^2 x - 2cos^2 x + psinx - pcosx = 0
p(sinx - cosx) = 2cos^2 x - 2sin^2 x
p (1/√3) = 2cos (2x)
p = 2√3cos(2x)

add the two original equations
2sin^2 x + 2cos^2 x + psinx + pcosx + 2q = 0
2(sin^2 x + cos^2 x) + p(sinx + cosx) + 2q = 0
2(1) + 2√3 cos(2x)(sinx + cosx) + 2q = 0
1 + √3(cosx - sinx)(cosx + sinx)(sinx + cosx) + q = 0
...... (since cos 2x = cos^2 x - sin^2 x, which is a difference of squares)

q = √3(sinx - cosx)(sin^2 x + 2sinxcosx + cos^2 x) - 1
= √3(1/√3)(1 + sin (2x) ) - 1
= sin 2x

finally:
p^2- 8q
= [2√3cos(2x)]^2 - 8sin(2x)
= 12 cos^2 (2x) - 8sin(2x)

check my work, so many places to make mistakes here.

I know I could have changed ...
p = 2√3cos(2x)
= 2√3(cos^2 x - sin^2 x)
= -2√3(sin^2 x - cos^2 x)
= -2√3(sinx - cosx)(sinx + cosx)
= -2√3(1/√3)(sinx + cosx)
= -2(sinx+cosx)

which would have given me
p^2 - 8q
= 4(sin^2 x + 2sinxcosx + cos^2 x) - 8 sin(2x)
= 4(1 + sin(2x) - 8sin(2x)
= 4 + 4sin(2x) - 8sin(2x)
= 4 - 4sin(2x)

Well, that was fun!