TITLE: Why can ALL quadratic equations be solved by the quadratic formula?
QUESTION [291 upvotes]: In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.
Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?

REPLY [2 votes]: A quadratic equation has two solutions, $R$ and $S$:
$$x^2 + Bx + C = (x-R)(x-S) $$
Developing the right-hand side:
$$ x^2 + Bx + C = x^2 - (R+S)x + RS $$
So, we derive two equalities: $-B = R + S$ and $C = RS$.
Key: from the first equality, we see the average of the two roots is equal to $-\dfrac{B}{2}$
Therefore, we are looking for two numbers of the form $-\dfrac{B}{2} \pm z$, with unknown $z$.
Now, from the second equality above, we then see that:
$$ \left(-\frac{B}{2} + z\right) \left(-\frac{B}{2} - z\right) = C $$
Developing the left-hand size, we get:
$$ \left(-\frac{B}{2}\right)^2 - z^2 = C $$
Rearranging, we get:
$$ z = \pm \sqrt{\frac{B^2}{4}-C} $$
Replacing this in the solution $-\dfrac{B}{2} \pm z$, we get
$$ -\frac{B}{2} \pm \sqrt{\frac{B^2}{4}-C} $$
which is the quadratic formula.
(solution from here)

This methods offers a different way to solve the problem. Instead of remembering the quadratic solution, do this:

Compute the general form of the solution, $-\dfrac{B}{2} \pm z$, for arbitrary $z$.
Since their product must be equal to $C$, now compute $\left(-\dfrac{B}{2}\right)^2 - z^2 = C$ and find $z$
Replace back into the formula in 1. Done.