Quadratic Equations

Quadratic Equations

Description

The formula you may familiar with:

\[ax^2 + bx + c = 0\]

And the solution is given:

\[x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{and} \quad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\]

But why how they mathematicians produced this solution formula?

Prof

The idea and work through are all from the book No bullshit guide to linear algebra. It’s very interesting

Ok let’s work through it.

\[ax^2 + bx + c = 0\] \[ax^2 + bx = -c\] \[x^2 + \frac{b}{a}x = -\frac{c}{a}\]

Now from the left of expression, we’ll use Complete the square method.

\[(x - h)^2 + k = x^2 + \frac{b}{a}x \tag{1}\] \[x^2 - 2hx + h^2 + k = x^2 + \frac{b}{a}x\]

Observe this:

\[-2hx = \frac{b}{a}x\] \[-2h = \frac{b}{a}\] \[h = -\frac{b}{2a}\]

so we got

\[(x + \frac{b}{2a})^2\]

which can be express to:

\[\left(x + \frac{b}{2a}\right)^2 = \left(x + \frac{b}{2a}\right)\left(x + \frac{b}{2a}\right) = x^2 + \frac{b}{2a}x + x\frac{b}{2a} + \frac{b^2}{4a^2} = x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\] \[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b^2}{4a^2}\right) = \left(x^2 + \frac{b}{a}x\right)\]

Continue where we left:

\[\left(x^2 + \frac{b}{a}x\right) = \left(-\frac{c}{a}\right)\] \[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b^2}{4a^2}\right) = \left(-\frac{c}{a}\right)\] \[\left(x + \frac{b}{2a}\right)^2 = \left(-\frac{c}{a}\right) + \left(\frac{b^2}{4a^2}\right)\] \[\left(x + \frac{b}{2a}\right) = \pm \sqrt{ \left(-\frac{c}{a}\right) + \left(\frac{b^2}{4a^2}\right)}\] \[\left(x + \frac{b}{2a}\right) = \pm \sqrt{ \left(-\frac{(4a)c}{(4a)a}\right) + \left(\frac{b^2}{4a^2}\right)}\] \[\left(x + \frac{b}{2a}\right) = \pm \sqrt{ \left(\frac{-4ac + b^2}{4a^2}\right)}\] \[\left(x + \frac{b}{2a}\right) = \pm \left(\frac{\sqrt{b^2 -4ac}}{2a}\right)\] \[x = \frac{-b}{2a} \pm \left(\frac{\sqrt{b^2 -4ac}}{2a}\right)\] \[x = \left(\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\right)\]