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

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Answer 1 · 2016-03-27T12:33:51

#x=0# or #x=1/4#

As #a^2+2ab+b^2=(a+b)^2#

Trying to convert #4x^2-x# in form #a^2+2ab# we can have

#4x^2-x=(2x)^2+2xx2x xx(-1/4)#

This means we should add #(-1/4)^2# on both sides for completing the square.

Hence, #4x^2-x=0# can be written as

#(2x)^2+2xx2x xx(-1/4)+(-1/4)^2-(-1/4)^2=0# or

#(2x-1/4)^2-(-1/4)^2=0# or

#(2x-1/4-1/4)(2x-1/4+1/4)=0# or

#(2x-1/2)*(2x)=0# or

#2(x-1/4)*2x=0# and dividing by #4# and rearranging we have

#x(x-1/4)=0# i.e. #x=0# or #x=1/4#