How do you solve the quadratic equation by completing the square: #x^2-4x+2=0#?

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Answer 1 · 2015-07-19T10:40:42

Complete the square to find #(x-2)^2 = x^2-4x+4 = 2#

Hence #x = 2 +-sqrt(2)#

Add #2# to both sides to get:

#2 = x^2-4x+4 = (x-2)^2#

So #x-2 = +-sqrt(2)#

Add #2# to both sides to get:

#x = 2 +-sqrt(2)#

In the general case:

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

from which we can derive the quadratic formula for solutions of #ax^2+bx+c = 0#:

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

Notice the #b/(2a)# term that gives us:

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