(a) (4m)x^2 - 2(m + n)x + n = 0 (b) (a + b + c)x^2 - 2(a + b) + (a + b - c) = 0 -- Prove that these equations have rational roots.

For roots of any quadratic equation to be rational, discriminant D must be a perfect square.

We know that for quadratic equation ax^2 + bx + c = 0, D = b^2 - 4ac

(a) For (4m)x^2 - 2(m + n)x + n = 0

D

= [-2(m + n)]^2 - 4(4m)(n)

= 4(m + n)^2 - 16mn

= 4(m^2 + n^2 + 2mn) - 16mn

= 4m^2 + 4n^2 + 8mn - 16mn

= 4m^2 + 4n^2 - 8mn

= 4(m^2 + n^2 - 2mn)

= 4(m - n)^2

= 2^2 . (m - n)^2

= (2m - 2n)^2 

which is a perfect square. Since discriminant of this equation is a perfect square, it will have rational roots.

(b) For(a + b + c)x^2 - 2(a+b) + (a + b - c) = 0

D

= [-2(a + b)]^2 - 4(a + b + c)(a + b - c)

= 4(a + b)^2 - 4(a^2 + ab - ac + ab + b^2 - bc + ac + bc - c^2)

= 4(a^2  + b^2 + 2ab) - 4(a^2 + b^2 - c^2 + 2ab)

= 4a^2 + 4b^2 + 8ab - 4a^2 - 4b^2 + 4c^2 - 8ab

= 4c^2

= (2c)^2

which is a perfect square. Since the determinant is a perfect square, the equation will have rational roots.