Roots of a Quadratic Equation

The key concept is substitution to verify an invariant root, followed by factorisation. Let f(x)=x^2−(a+6)x+6a. On substituting x=6, f(6)=36−6(a+6)+6a=36−6a−36+6a=0 for every value of a. Hence 6 is always a root. Indeed, the equation factors as (x−6)(x−a)=0, so the second root is a. Statement A is not generally true: a+6 is not a root for arbitrary a. The roots are equal only in the special case a=6, and for every real a the roots 6 and a are real, so C and D are false. Therefore B is the only statement that is universally valid.