Deriving the quadratic formula without completing the square

Consider the general quadratic equation

ax^2+bx + c =0, a \ne 0.\quad\quad\quad(*)

Let

y = x + \omega \implies x = y-\omega.\quad\quad\quad(1)

Substituting (1) into (*), we have

a(y-\omega)^2+b(y-\omega)+c = 0.

This is a new quadratic equation in y:

ay^2+(b-2a\omega)\cdot y + a\omega^2-b\omega+c=0.\quad\quad\quad(2)

Let

\omega = \frac{b}{2a} \quad\quad\quad(3)

so that (2) becomes

ay^2 + 0\cdot y + a(\frac{b}{2a})^2 - b(\frac{b}{2a}) +c=0.

That is,

ay^2  + \frac{-b^2}{4a} + c =0.\quad\quad\quad(4)

Solving (4) for y^2,

y^2 = \frac{b^2-4ac}{4a^2} \implies y = \pm \frac{\sqrt{b^2-4ac}}{2a}\quad\quad\quad(5)

Substituting (3) and (5) into (1) gives

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

i.e.,

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

4 thoughts on “Deriving the quadratic formula without completing the square

  1. ax^2 + bx + c = 0
    multiply by 4a:
    4a^2x^2 + 4abx + 4ac = 0
    add b^2 to each side and move 4ac:
    4a^2x^2 + 4abx + b^2 = b^2 – 4ac
    recognize as a perfect square:
    (2ax + b)^2 = b^2 – 4ac
    take square roots:
    2ax + b = +- sqrt(b^2 – 4ac)
    solve for x:
    x = (-b +- sgrt(b^2 – 4ac)) / 2a

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