What is the quadratic formula?

1 Answer
Mar 2, 2017

Answer:

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

Explanation:

Given a quadratic equation in the standard form:

#ax^2+bx+c = 0#

the roots are given by the quadratic formula:

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

This is a very useful formula to memorise, but where does it come from?

Given:

#ax^2+bx+c = 0#

Note that:

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

#color(white)(a(x+b/(2a))^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#

#color(white)(a(x+b/(2a))^2) = (b^2-4ac)/(4a)#

Divide both sides by #a# to get:

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

#color(white)((x+b/(2a))^2) = (b^2-4ac)/(2a)^2#

Take the square root of both sides, allowing for both positive and negative square roots:

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

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

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

Note that the quadratic formula will always work, but it will only give real values if #b^2-4ac >= 0#, resulting in real square roots.