Deriving the Quadratic Formula
This is another instance of a “well-known derivation”. Go to any search engine and enter Deriving the Quadratic Formula by completing the square. You will find a few videos on YouTube that follow a similar process.
The Quadratic Formula is probably one of the most well-known formulae from mathematic (namely algebra). I still find myself saying
negative “b” plus or minus the square root of “b-squared” minus four ‘a’ ‘c’ over twice ‘a’
as a pneumonic device. However, until recently I never worked through a complete derivation.
Our goal is to derive:
$$ \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} $$
From our math foundations, we know the form of a degree-2 polynomial:
$$ \begin{array}{rl} \varphi &=& \sum\limits_{i=0}^{2} c_{i}x^{i} \\ &=& c_0 + c_{1}x +c_{2}x^{2} \\ &=& c + bx + ax^2 \\ &=& ax^2 + bx + c \\ \end{array} $$
Take note of the notation. We will use $\sum$, $c_i$ and $\varphi$ throughout the semester… it is just notation.
The first step is to set $\varphi$ equal to zero:
$$ \begin{array}{rl} \varphi &=& 0 \\ ax^2 + bx + c &=& 0 \\ a(x^2 + \frac{bx}{a}) + c &=& 0 \\ a(x^2 + \frac{bx}{a}) + c &=& 0 \\ \end{array} $$
This form allows us to complete the square:
$$ \begin{array}{rl} a(x^2 + \frac{bx}{a}) + c &=& 0 \\ a(x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}) + c &=& 0 \\ a(x^2 + \frac{bx}{a} + \frac{b^2}{4a^2}) - \frac{b^2}{4a} + c &=& 0 \\ \end{array} $$
After completing the square, the rest is algebraic manipulation.
$$ \begin{array}{rl} a(x^2 + \frac{bx}{a} + \frac{b^2}{4a^2}) - \frac{b^2}{4a} + c &=& 0 \\ a(x^2 + \frac{bx}{a} + \frac{b^2}{4a^2}) &=& \frac{b^2}{4a} - c \\ a(x + \frac{b}{2a})^2 &=& \frac{b^2}{4a} - c \\ a(x + \frac{b}{2a})^2 &=& \frac{b^2}{4a} - \frac{4ac}{4a} \\ a(x + \frac{b}{2a})^2 &=& \frac{b^2 - 4ac}{4a} \\ (x + \frac{b}{2a})^2 &=& \frac{b^2 - 4ac}{4a^2} \\ \sqrt{(x + \frac{b}{2a})^2} &=& \pm \sqrt{\frac{b^2 - 4ac}{4a^2} } \\ \sqrt{(x + \frac{b}{2a})^2} &=& \pm \frac{\sqrt{b^2 - 4ac} }{\sqrt{4a^2} } \\ (x + \frac{b}{2a}) &=& \frac{\pm \sqrt{b^2 - 4ac} }{2a} \\ x + \frac{b}{2a} &=& \frac{\pm\sqrt{b^2 - 4ac} }{2a} \\ x &=& \frac{-b}{2a} \frac{\pm\sqrt{b^2 - 4ac} }{2a} \\ x &=& \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} \\ \end{array} $$
This should show why analytically solving a problem is better than plugging in numbers, first. We now have a general solution (one you have utilized quite frequently). We will take a similar approach throughout this semester (i.e., we will favour solving the general problem, and defer plugging in numbers).