0%

Being tired of memorizing this fomula? Let’s take a look at it from a different perspective.
$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ \$?

Consider $ax^2+bx+c=0$, and to normalize the equation, provided that $b’=b/a$ and $c’=c/a$, then we have $x^2+b’x+c’=0$.

Suppose we have r and s, such that $x^2+b’x+c’=(x-r)(x-s)=0$, then it is easy to see that: $b’=-(r+s)$ and $c’=rs$.

Now, think of the middle point m and the deviation d, where $r=m-d$ and $s=m+d$, as show in below snapshot:

Then we can re-write $b’$ and $c’$ as $b’=-2m$ and $c’=(m+d)(m-d)=m^2-d^2$, thus, $m=-b’/2$ and $d^2=m^2-c’$.

Meanwhile, the roots can be re-written as $x=m \pm d$, where $m=-b’/2$ and $d=\sqrt{m^2-c’}$.

For example, if $x^2-21x+110=0$, from above we have $m=10.5$ and $d=\sqrt{10.5^2-110}=0.5$, so the roots are $x=10.5\pm 0.5$, i.e. 10 and 11.

These are just notes I took from the amazing videos from 3Blue1Brown. I highly suggest everyone who’s interested in details watch it. The link is down below.

Reference: