How do you solve #2x^2 - 8x + 16 = 0# by completing the square?

1 Answer
Shwetank Mauria
Apr 10, 2016

#x=2-2i# or #x=2+2i#

Explanation:

Each monomial in equation #2x^2-8x+16=0#, is divisible by #2#, so diving by #2#, we get

#x^2-4x+8=0#

Now comparing it with #(x-a)^2=x^2-2ax+a^2#, it is apparent that if #-2a=-4#, we have #a=2# and #a^2=4#.

Hence adding #4#, we can complete the square. Doing so gives us

#x^2-4x+4-4+8=0# or

#(x-2)^2+4=0# but as in the domain of real numbers the LHS of the equation will be always positive, we cannot have real roots but only complex roots. Hence we write #+4# as #-(2i)^2 and then equation becomes

#(x-2)^2-(2i)^2=0# or #(x-2+2i)(x-2-2i)=0#

i.e. #x=2-2i# or #x=2+2i#

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