1. ## Deriving the Quadratic Formula

Have you ever wondered how the quadratic formula was derived? Well, if your a proof-addict like I am, or perhaps just someone that clicked on the wrong link,but now has become a little curious as to what the quadratic formula is all about, let me put you at ease.

Definition:

Quadratic Equation-A polynomial equation of degree 2

Degree 2 means that the highest power of the independent variable is 2 ( e.g $x^2$ is of degree 2, $x^3$ is of degree 3, and etc...)

Here is the quadratic equation in standard form:

$ax^2+bx+c=0$

It is important to understand that we call this the "standard" form because it is "standard" practice to move all terms to the left side of the equation such that the right side of the equation is zero. The word standard is not used arbitrarily, having the right side equal to 0 allows us to easily solve what would ordinarily be very difficult, if not impossible, equations.

For example, having the equation equal to zero allows us to swiftly find roots by factoring:

$x^2-2x+3=0\Longleftrightarrow{(x-3)(x+1)=0}\Rightarrow{x=-1,3}$

But what about $6x^2+28x-80=0$

I don't know about you, but I don't want to waste my time trying to find all of the factors whose product is -80, and whose sum is +28. There's an easier way! The quadratic formula!

So, here is how it was derived.

$ax^2+bx+c=0$

Divide Both Sides by $a$.

$\frac{ax^2}{a}+\frac{b}{a}x+\frac{c}{a}=\frac{0}{a }$

$=x^2+\frac{b}{a}x+\frac{c}{a}=0$

Now Subract both sides by $\frac{c}{a}$

$=x^2+\frac{b}{a}x+\frac{c}{a}-\frac{c}{a}=0-\frac{c}{a}$

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

Here is where the ingenuity came in to play. Add $\frac{1}{2}$ of the of the coefficient squared of the linear term to both sides (the linear term is the term with the variable raised to the first power).This algebraic maneuver is known as "completing the square".

$=x^2+bx+(\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}$

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

Simplifying Yields (Finding a Common Denominator)

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

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

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

Now, here we recognize that the left side of the equation can be factored, because it is a trinomial square.

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

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

Now taking the square root

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

And finally subtracting $\frac{b}{2a}$ from both sides and simplifying

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

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

The last step is justified by the fact that $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

I'm excited. Isn't that sheer brilliance. Sure, we can all substitute to find ways around problems, but can we take it to the next level? Can we synthesize? I only hope to be able to thinoutside of the box like this some day.

By the way, the problem of solving quadratic equations plagued humanity for thousands of years before this formula was discovered. So all all you guys/gals in Abstract Algebra and Differential Equations, take a moment to appreciate the efforts of your forebears, without whom, we'd all still be trying to figure out how to solve right triangles. Oh! Now there was a genious! Good old Pythagoras. How did he show..............

BBC - h2g2 - The History Behind The Quadratic Formula

2. Originally Posted by VonNemo19
Now, here we recognize that the left side of the equation can be factored, because it is a trinomial square.

$(x+\frac{b}{2})(x+\frac{b}{2})=\frac{b^2-4ac}{4a^2}$
A small correction: It should read $(x+\frac{b}{2a})(x+\frac{b}{2a})=\frac{b^2-4ac}{4a^2}$. And a few lines that follow have the same typo...

3. thanks man, that was a biggie.

4. Below, I will write a solution for the roots of the quadratic formula, which I figured out myself when I was in first or second class in the university.
Consider
$ax^{2}+bx+c=0$ for $x\in\mathbb{R}$.....(1)
If $a^{2}+c^{2}=0$, the solution is very simple, hence we consider the case $ac\neq0$ below.
Since $c\neq0$, we know that $x=0$ is not a solution of (1).
Hence we may divide (1) by $x$ after taking $b$ to the right-hand side to get
$ax+\frac{c}{x}=-b$.....(2) (the trick here is that the middle term after squaring destroys the unknown, hence we may push it to the other side of the equation to leave the unknowns on one side)
Squaring (2), we get
$a^{2}x^{2}+2ac+\frac{c^{2}}{x^{2}}=b^{2}$.....(3)
Now add $-4ac$ on both sides of (3) to get
$a^{2}x^{2}-2ac+\frac{c^{2}}{x^{2}}=b^{2}-4ac$.....(4)
Taking now square roots of both sides of (4), we get
$ax-\frac{c}{x}=\pm\sqrt{b^{2}-4ac}$.....(5)
Finally, sum (2) and (5) to get
$2ax=-b\pm\sqrt{b^{2}-4ac}$
or equivalently
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$....(6)

Note. The formula (6) also holds for the case $c=0$.

5. bkarpuz, another derivation of the QUADRATIC FORMULA:

(1) ax^2 + bx + c = 0, multiply BS by 4a,

(2) 4a(x)^2 + 4abx = -4ac, regroup them,

(3) (2ax)^2 + 2(2ax)b = -4ac, add b^2 to BS,

(4) (2ax)^2 + 2(2ax)b + (b)^2 = b^2 - 4ac. Well, LHS is a perfect trinomial,

(5) (2ax + b)^2 = b^2 - 4ac, take square root of BS,

(6) 2ax + b = (+/-)(b^2 - 4ac)^(1/2), transpose b to the RHS,

(7) 2ax = -b (+/-)(b^2 - 4ac)^(1/2), DBS by 2a, thus

(8) x = -b (+/-)(b^2 - 4ac)^(1/2) / 2a