If `a+b+c=0` and `a,b,c` are rational. Prove that the roots of the equation
`(b+c-a)x^(2)+(c+a-b)x+(a+b-c)=0` are rational.

Explanation

Given equation is
`(b+c-a)x^(2)+(c+a-b)+(a+b-c)=0`………i
`:'(b+c-a)+(c+a-b)x+(a+b-c)=a+b+c=0`
`:.x=1` is a root of Eq. (i) let other root of Eq. (i) is `alpha` then
Product of roots`=(a+b-c)/(b+c-a)`
`implies1xxalpha=(-c-c)/(-a-a)[:'a+b+c=0]`
`:.alpha=c/a`[rational]`
Hence, both roots of Eq. (i) are rational.
Aliter
Let `b+c-a=A,c+a-b=B,a+b-c=C`
Then `A+B+C=0 [ :' a+b+c=0]`.........ii
Now Eq. (i) becomes
`Ax^(2)+BX+C=0`..........iii
Discriminant of Eq. (iii)
`D=B^(2)-4AC`
`=(-C-A)^(2)-4AC[ :'A+B+C=0]`
`=(C+A)^(2)-4AC`
`=(C-A)^(2)-(2a-2c)^(2)`
`=4(a-c)^(2)=A` perfect square
Hence, roots of Eq. (i) are rational.