How do you solve #4x^2-8x + 1 = 0#?

#4x^2-8x + 1 = 0#

#4x^2-8x + 1 = 0|:4#
#x^2-2x+1-1+1/4=0#
#(x-1)^2-1+1/4=0#
#(x-1)^2-3/4=0|+3/4#
#(x-1)^2=3/4|sqrt()#
#x-1=+-sqrt(3)/2|+1#
#x=1+-sqrt(3)/2#

Completing the square.

Given: #y=0=4x^2-8x+1#

Write as:
#y=0=4(x^2-8/4x)+1#

We now change this into the format of completing the square but in doing so we introduce a value that is not in the original equation. We 'get rid' of this by introducing a value that changes it into 0. Let this as yet unknown value be #k#

#0=4(x^2color(white)("d")ubrace(-8/4)color(white)("d")x)+1#
#color(white)("dddddddd.d")darr#
#color(white)("ddddddd")"Halve this"#
#color(white)("dddddddd.d")darr#
#0=color(white)("d")4(xcolor(white)("d.")obrace(-1)color(white)("d"))^2+k+1#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set #4(-1)^2+k=0 color(white)("dddd")=>color(white)("dddd")k=-4#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#0=4(x-1)^2-4+1#

#0=4(x-1)^2-3#

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

#x-1=+-sqrt(3)/2#

#x=1+-sqrt(3)/2# Exact values

#x~~0.13397...#
#x~~1.86602...#

#x~~0.13# to 2 decimal places
#x~~1.87# to 2 decimal places

#4x^2-8x+1=0#

Using the quadratic formula

#:.ax^2+bx+c=0#

#:.a=4,b=-8,c=1#

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

#:.x=(-(-8)+-sqrt((-8)^2-4(4)(1)))/(2(4))#

#:.x=(8+-sqrt 48)/8#

#:.x=(sqrt(4*4*3))/8#

#:.x=(8+-4 sqrt 3)/8#

#:.x=cancel 8^1/cancel 8^1+-(cancel4^1 sqrt 3)/cancel 8^2#

#:.x=1+-sqrt 3/2#

#:.x=1+sqrt 3/2 or x=1-sqrt 3/2#