# The Quadratic Formula
$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ ^l9kegs
## Simpler Version
$
\begin{gather}
ax^2 + bx + c = 0 & \text{divide both sides by $a$} \\
x^2 + b'x + c' = 0 & b' = \frac{b}{a} \quad c'= \frac{c}{a} \\
m = -\frac{b'}{2} & \text{midpoint between two solutions} \\
x_{1,2} = m \pm \sqrt{m^2-c'} \\
\end{gather}
$
^5vr6d5
### Proof for the Simpler Version
Lets define the midpoint of the two solutions as $m=\frac{x_1 + x_2}{2}$, and a value $d$ that represents how far away the two solutions are from $m$. Then, the product of the solutions can be written as:
$
\begin{align}
x_1x_2 &= (m-d)(m+d) \\
x_1x_2 &= m^2-d^2 \\
d^2 &= m^2 - x_1x_2 \\
d &= \sqrt{m^2-x_1x_2}
\end{align}
$
As $x_{1,2} = m \pm d$ and we defined d, all we have to do now is figure out the product and $m$ values and we're done.
Lets go back to the standard form of the quadratic equation and turn it into a [[Polynomials#Monic Polynomial]] by dividing both sides by $a$.
$
\begin{align}
ax^2 + bx + c &= 0 \\
x^2 + b'x + c' &= 0 && b' = \frac{b}{a} && c'= \frac{c}{a}
\end{align}
$
As per [[Vieta's Formulas#For Quadratic -s|Vieta's Formulas]], $x_1+x_2 = -b'$ and $x_1x_2 = c'$. From there $m=\frac{-b'}{2}$.
#### You Can Represent Any Product as Difference of Squares
While explaining the simpler version, 3Blue1Brown[^1] points out that we can represent any product of two numbers as the difference of squares: namely the square of their mean, and the square of the difference from the mean:
$rs = m^2 - d^2$
where $m = \frac {r+s} 2$ and $d = \sqrt {m^2 - rs}$.
## Proof By Completing The Square
$
\begin{align}
ax^2+bx+c &= 0 \\
ax^2 +bx &= -c && \text{divide both sides by $a$} \\
x^2 + \frac b ax &= -\frac c a && \text{add $\left(\frac b {2a}\right)^2$ to both sides} \\
x^2 + \frac b ax + \left(\frac b {2a}\right)^2 &= \left(\frac b {2a}\right)^2 -\frac c a \\
\left(x + \frac b {2a}\right)^2 &= \frac {b^2} {4a^2} -\frac c a \\
\left(x + \frac b {2a}\right)^2 &=\frac {b^2-4ac} {4a^2} \\
x + \frac b {2a} &= \pm \sqrt{\frac {b^2-4ac} {4a^2}} \\
x &= -\frac b {2a} \pm \frac {\sqrt{b^2-4ac}} {\sqrt{4a^2}} \\
x &= -\frac b {2a} \pm \frac {\sqrt{b^2-4ac}} {2a} \\
x &= \frac {-b \pm \sqrt{b^2-4ac}} {2a}
\end{align}
$
[^1]: https://www.youtube.com/watch?v=MHXO86wKeDY