Example 1 If $x_{0}$ is a root of the quadratic equation $a x^{2}+b x+$ $c=0(a \neq 0)$, then the relationship between the discriminant $\Delta=b^{2}-4 a c$ and the quadratic form $M=\left(2 a x_{0}+b\right)^{2}$ is ( ).
(A) $\Delta>M$
(B) $\Delta=M$
(C) $\Delta<M$
(D) cannot be determined
(1992, National Junior High School Mathematics League)
There are many methods to solve this problem, but no matter which method is used, the final result we emphasize (i.e., equation (2)) must be obtained: the discriminant is not an isolated $b^{2}-4 a c$, it is related to the quadratic equation and is expressed as a perfect square. The solution method that best reflects the proposition's intent is:

Solution 1: Completing the square from (1) to (2) (omitted).

Solution 2: From the quadratic formula, we have
$$
x_{0}=\frac{-b \pm \sqrt{\Delta}}{2 a} \text {, }
$$

which implies $2 a x_{0}+b= \pm \sqrt{\Delta}$. Squaring both sides gives $\left(2 a x_{0}+b\right)^{2}=\Delta$, hence the answer is (B).

Solution 1 provides a new method for completing the square for a quadratic equation, while Solution 2 uses the analytical method (returning from the quadratic formula to the quadratic equation) to find a new way to complete the square. The benefits of this new method are:

1. It is simpler than the original method of dividing both sides by $a$.
2. It effectively eliminates the possible misunderstanding $\sqrt{a^{2}}=a$ in the original derivation:
$$
x+\frac{b}{2 a}= \pm \sqrt{\frac{b^{2}-4 a c}{4 a^{2}}}= \pm \frac{\sqrt{b^{2}-4 a c}}{2 a}
$$
3. It highlights the origin, essence, and importance of the discriminant.
$(2 a x+b)^{2}=b^{2}-4 a c$ clearly emphasizes that the discriminant originates from the method of completing the square, is the result of completing the square, and is essentially a perfect square $(2 a x+b)^{2}$.

1.3 A Troubled and Eventful Early Competition Problem
In 1956, China held its first mathematics competition, and Beijing posed the following problem involving the discriminant.