# Deriving the quadratic formula without completing the square $ax^2+bx + c =0, a \ne 0.\quad\quad\quad(*)$

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. Johnny Weilharter

Reblogged this on Mathematik mit CAS Maxima und Geogebra and commented:
Michael Xue ist ein CAS-Maxima Profi.

2. Vin Underwood

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

• Michael Xue

Thank you for showing another way of completing the square.

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