You can put this solution on YOUR website! if a+b+c=0 and a,b,c are rational then the roots of the equation
(b+c-a)x²+(c+a-b)x+(a+b-c)=0 are
That's
[(b+c)-a]x² + [(c+a)-b]x + [(a+b)-c] = 0
Solve a+b+c = 0 for (b+c), (c+a), and (a+b)
b+c = -a
c+a = -b
a+b = -c
Substituting these into
[(b+c)-a]x² + [(c+a)-b]x + [(a+b)-c] = 0
[-a-a]x² + [-b-b]x + [-c-c] = 0
-2ax² - 2bx -2c = 0
ax² + bx + c = 0
That's just the general quadratic equation. So its roots are given
by the quadratic formula:
Edwin
