SOLUTION: 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

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Question 631393: 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
Answer by AnlytcPhil(1810)   (Show Source): 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
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