How do you solve #x^ { 2} - x - 1= ( x - \frac { 1} { 2} ) ^ { 2} - \frac { 5} { 4}#?

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Answer 1 · 2017-02-02T20:56:49

#oo# solutions

First, you would expand the square #(x-1/2)^2#, then:

#cancelx^2cancel(-x)-1=cancelx^2cancel(-x)+1/4-5/4#

#-1=-1#

that is an identity and there are #oo# solutions

Answer 2 · 2017-02-02T21:31:15

All values of #x# are solutions to this equation.

#x^2-x-1=(x-1/2)^2-5/4#

Expand the quadratic on the right hand side.

#x^2-x-1=x^2-x+1/4-5/4#

Combining like terms, we find that

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

But, of course, this just means that #x=x#, which is true for every single possible value of #x#!