how do prove that this equation has rational roots for all rational values of K 3x^2+kx-k=3x

Answer 1

first write your equation in standard quadratic form

3x^2 + kx - 3x - k = 0
3x^2 + (k-3)x - k = 0
so
a=3
b=k-3
c=-k

the discriminant is b^2 - 4ac
= (k-3)^2 - 4(3)(-k)
= k^2 - 6k + 9 + 12k
= k^2 + 6k + 9
= (k+3)^2

for rational roots, (k+3)^2 ≥ 0
but the square of anything is always ≥ 0
so your equation has rational roots for any value of k
(no matter what you put in for k, (k+3)^2 is > 0 )