Question

How do you derive the quadratic formula?

Answer 1

See explanation...

Explanation:

Given `ax^2+bx+c = 0` to solve.

Note that:

`a(x+b/(2a))^2=a(x^2+b/ax+b^2/(4a^2))=ax^2+bx+b^2/(4a)`

So:

`0 = ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))`

Add `b^2/(4a)-c` to both ends and transpose to get:

`a(x+b/(2a))^2 = b^2/(4a) - c=(b^2-4ac)/(4a)`

Divide both sides by `a` to get:

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

Take the square root of both sides (allowing for either sign) to get:

`x+b/(2a) = +-sqrt((b^2-4ac)/(4a^2)) =+-sqrt(b^2-4ac)/(2a)`

Subtract `b/(2a)` from both sides to get:

`x = -b/(2a)+-sqrt(b^2-4ac)/(2a) = (-b+-sqrt(b^2-4ac))/(2a)`

Note that all of this is based on simple properties of arithmetic, so will work regardless of whether `a`, `b` and `c` are integers, rational, irrational or even Complex numbers.