If `aa n dc` are odd prime numbers and `a x 62+b x+c=0` has rational roots , where `b in I ,` prove that one root of the equation will be independent of `a ,b ,cdot`

Explanation

Consider the equation
`ax^(2) + bx + c = 0`
Since tha roots rational, discriminant of the given quadratic equation will be a perfect square. Hence,
`b^(2) - 4ac = lamda^(2), lambda in I`
`b^(2) - lamda^(2)= 4 ac or (b+lambda)(b-lambda) =4ac `
Then, we have the following possibilities:
`b + lambda = 2a, b - lambda = 2c`
or `b+ lambda = 2c, b - lambda = 2a`
or `b+ lambda = - 2a, b - lambda = - 2c`
or `b + lambda = - 2c, b - lambda = - 2a`
(as a + `lambda and b - lambda` bothe should be even)
Solving the above cases, we get
`b = ne (a + c), lambda = pm (a - c)`
Hence, the roots are `-(b - lambda)//2a`. Clearly, one of the roots is `-1//2` (Put the value of `lambda` and simplify to get ` -1//2).