How do I solve #4x^2 - 6x + 1 = 0# by completing the square?

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Answer 1 · 2014-09-13T22:54:49

#4x^2-6x=-1#

Factor out a #4#

#4(x^2-6/4x)=-1#

Simplify fraction

#4(x^2-3/2x)=-1#

Take half of x term and square the result

#((-3/2)/2)^2=((-3/2)*(1/2))^2=(-3/4)^2=9/16#

Add #4*9/16# or #9/4# to the right side because we factored out a #4#.
Add #9/16# to the left side.

#4(x^2-3/2x+9/16)=-1+9/4#

#4(x-3/4)^2=-1+9/4#

#4(x-3/4)^2=-4/4+9/4#, common denominators #-1=-4/4#

#4(x-3/4)^2=5/4#

#(x-3/4)^2=(5/4)/4#

#(x-3/4)^2=5/16#

#sqrt((x-3/4)^2)=+-sqrt(5/16)#

#x-3/4=+-sqrt(5/16)#

#x-3/4=+-sqrt(5)/4#

#x=+-sqrt(5)/4+3/4#

#x=(+-sqrt(5)+3)/4#