How do you solve #x^2 - 4x +2 = 0# using completing the square?

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Answer 1 · 2015-06-11T22:00:11

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

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

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

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

To solve #(x-2)^2-2 = 0#, first add #2# to both sides to get:

#(x-2)^2 = 2#

Then #(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 deduce the quadratic formula:

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