Bunuel wrote:

Which of the following equations has a root in common with x^2−7x+12=0 ?

A. x^2+1=0

B. x^2−x−12=0

C. x^2−8x−4=0

D. 3x^2−9=0

E. x^2+x-6=0

If we carefully look at the given equation we can arrange it in the following manner:

(x-3)(x-4) = 0

so the two roots are 3 and 4.

Now put 3 and 4 in given equations. The equation in which one of them gives value 0, that will be our answer.

(A) putting 3: we get 10; Putting 4 we get 17. Reject this option [by the way x^2+1 cannot have real roots]

(B) putting 3: we get 6; Putting 4 we get 0. This is the equation.

We are lucky . no need to check other options.

B is the answer.