Quote Originally Posted by HallsofIvy View Post
No, it's not. "0= 0" is true for any values of A and B.

If both A and B satisfy $\displaystyle x^2+ Ax+ B= 0$ then $\displaystyle A^2+ A(A)+ B= 2A^2+ B= 0$, so that $\displaystyle B= -A^2$, and $\displaystyle B^2+ AB+ B= 0$. Since $\displaystyle B= -A^2$, $\displaystyle (-A^2)^2+ A(-A^2)+ (-A^2)= A^4- A^3- A^2= A^2(A^2- A- 1)= 0$.