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

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Answer 1 · 2015-10-10T06:26:55

The solutions are:
#color(blue)(x= (1+sqrt(3))/2#

#color(blue)(x= (1-sqrt(3))/2#

#2x^2-2x-1=0#

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

The Discriminant is given by:
#Delta=b^2-4*a*c#
# = (-2)^2-(4*2*(-1))#
# = 4+8= 12#

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

#x = (-(-2)+-sqrt(12))/(2*2) = (2+-2sqrt(3))/4#

# (2+-2sqrt(3))/4= (cancel2(1+-sqrt(3)))/cancel4#

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

The solutions are:
#color(blue)(x= (1+sqrt(3))/2#

#color(blue)(x= (1-sqrt(3))/2#