Deriving the Quadratic Formula (English)

I will show alternative way to accomplish this task below.

You can optionally see https://www.youtube.com/watch?v=OZNHYZXbLY8 and http://www.youtube.com/watch?v=bjH1HphOZ1Y for what is completing square. You can do it another way by using formula

There is more below.
Ниже есть продолжение.

$(a+b)^2=a^2+2ab+b^2$

So the tough question will be

$x^2+3x+4=2$

$x^2+2*\frac{3}{2}x+4=2$

$x^2+2*\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+4=2$

$(x+\frac{3}{2})^2=2-4+\frac{9}{4}$

$(x+\frac{3}{2})^2=\frac{1}{4}$

$x+\frac{3}{2}=\frac{1}{2}$ or $x+\frac{3}{2}=-\frac{1}{2}$

$x=-1$ or $x=-2$

And so, alternative derivation would be:

We want to solve

$ax^2+bx+c=0$, where $a\ne0$

Because $a\ne0$ we can divide both parts by a, we will get

$0=x^2+\frac{b}{a}x+\frac{c}{a}=x^2+2\frac{b}{2a}x+\frac{c}{a}=x^2+2\frac{b}{2a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}=$
$=(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a}$

or another way around

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

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

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

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

From here:

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

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

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$


After, you know the formula, solving the tough question will be much easier:


$x^2+3x+4=2$

$x^2+3x+2=0$

$x_{1,2}=\frac{-3\pm\sqrt{3^2-4*1*2}}{2*1}=\frac{-3\pm\sqrt{1}}{2}=\frac{-3\pm 1}{2}$

$x_{1}=\frac{-3+1}{2}=-1$

$x_{2}=\frac{-3-1}{2}=-2$