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

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Answer 1 · 2015-08-21T07:45:03

The solutions are:
#x=color(blue)((1-sqrt3)/2#

#x=color(blue)((1+sqrt3)/2#

#2x^2−2x=1#

#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= (2(1+-sqrt3))/4#

#=(cancel2(1+-sqrt3))/cancel4#

#=(1+-sqrt3)/2#

The solutions are:
#x=color(blue)((1-sqrt3)/2#
#x=color(blue)((1+sqrt3)/2#