How do you derive the quadratic formula?

1 Answer
Feb 17, 2016

Answer:

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.