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

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Answer 1 · 2015-08-21T07:36:49

The solutions are:
#color(blue)(x=(2+sqrt(2))/2#
#color(blue)(x=(2-sqrt(2))/2#

#2x^2-4x+1#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=2, b=-5, c=1#

The Discriminant is given by:
#Delta=b^2-4*a*c#
# = (-4)^2-(4*2*1)#
# = 16-8=8#

The solutions are found using the formula:
#x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-4)+-sqrt(8))/(2*2) = (4+-2sqrt(2))/4#

#=(2 (2+-sqrt(2)))/4 = (2+-sqrt(2))/2#

The solutions are:

#color(blue)(x=(2+sqrt(2))/2#
#color(blue)(x=(2-sqrt(2))/2#