Ax^2+bx=bx^2+a prove that it's roots are rational for all real values of a and b
ax^2+bx=bx^2+a
(a-b)x^2 + bx - a = 0
The discriminant is
b^2 + 4a(a-b) = b^2-4ab+4a^2 = (b-2a)^2
since that is a perfect square, plugging it into the quadratic formula will always yield a rational result if a and b are rational.
Note that the question is invalid if a or b is not rational. General real values will not work. If, say,
a=π and b=∛7
the roots will not be rational.
Also, it's its in this case, not it's.