How was the quadratic formula created? I have tested the quadratic formula and I have found that it works, yet I am curious as to how it was created. Can anybody please tell me one of the ways that it was created?
We begin with the equation $ax^2 + bx + c = 0$, for which we want to find $x$. We divide through by $a$ first, and then bring the constant term to the other side:

$$x^2 + \frac b a x = - \frac c a $$

Next, we complete the square on the left-hand side. Remember, to complete the square, you take half of the coefficient of the linear term, square it, and add it both sides. This means we add $(b/2a)^2 = b^2/4a^2$ to both sides:

$$x^2 + \frac b a x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac c a$$

The left-hand side factors as a result, and we combine the terms on the right-hand side by getting a common denominator:

$$\left(x + \frac b {2a} \right)^2  = \frac{b^2 - 4ac}{4a^2}$$

We now take the square root of both sides:

$$x + \frac b {2a}   = \pm \sqrt{ \frac{b^2 - 4ac}{4a^2}}$$

Solve for $x$:

$$x =- \frac b {2a} \pm \sqrt{ \frac{b^2 - 4ac}{4a^2}}$$

Recall that $\sqrt{a/b} = \sqrt a / \sqrt b ^{ \; \text{(note 1)}}$. Using this property, the denominator of our root becomes $2a$, giving a common denominator with $-b/2a$. Thus,

$$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$$

yielding the quadratic formula we all know and love. $^{ \; \text{(note 2)}}$



Footnotes:



Note $(1)$ - The usual properties for roots, and exponents in turn (since $\sqrt[n] a = a^{1/n}$) that most people are familiar with, do not always hold. In particular, for example

$$\sqrt{\frac a b} = \frac{\sqrt a}{\sqrt b} \;\;\;\;\; \sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$$

do not hold if $a,b$ are complex numbers. They hold if $a,b$ are nonnegative real numbers (and, in the first property shown, $b \ne 0$). A well-known example of why the second does not hold involves $i$, the complex number such that $i^2 = -1$. If this property held,

$$-1 = i^2 = \sqrt{-1} \sqrt{-1} = \sqrt{(-1)\cdot (-1)} = \sqrt{1} = 1$$

but $-1 \ne 1$.

In light of this note, note that we do not necessarily have any problems splitting up the root as we do in the proof. Even if $a&lt;0$, $a^2 &gt; 0$ as a result (of course, we also assume $a,b,c$ are real numbers in this derivation).



Note $(2)$ - As a note of interest, you are only guaranteed that $x$ is a real solution whenever the discriminant of the quadratic is nonnegative. The discriminant is the expression under the root; thus

$$ax^2+bx+c = 0 \; \text{has real solutions if and only if} \; b^2-4ac \ge 0$$

In particular, if $b^2 - 4ac = 0$, then the solutions $x$ gives are the same (what is called a "double root" or a "root of multiplicity $2$"). For $b^2 - 4ac &gt;0$, you are ensured two distinct real values for your solutions.

If $b^2 - 4ac &lt; 0$, then your solutions will instead be complex numbers. You will still have two distinct solutions however.
The way to get it is actually a beautiful one. The original proof is different since algebra was still in an early stage of development, but here's how I would do it today:

Let's consider the equation $ax^2+bx+c=0$ If a=0, we have a first degree equation. Otherwise, our equation can be written as $x^2+\frac{b}{a}x +\frac{c}{a}=0$

On the other hand, $(x+\frac{b}{2a})^2=x^2+\frac{b}{a}x+\frac{b^2}{4a^2}$. This looks like an idea completely out of the air, but you can get it by trying different ones and checking which one works. As you can see, the first two terms look like those form the equation, so we can say that $x^2+\frac{b}{a}x = (x+\frac{b}{2a})^2 - \frac{b^2}{4a^2}$

Now go back to the equation and replace $x^2+\frac{b}{a}x$ We get that $(x+\frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0$ From this equation, it's just basic algebra.

$(x+\frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a}$ Take the square root on both sides and enjoy your formula (unless I made a mistake, which I probably have)