How do you solve #sqrt x = x - 6#?

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Answer 1 · 2016-07-22T23:25:27

From the relation #sqrtx=x-6# we have that #x>=0# and

#x-6>=0=>x>=6#.Finally solutions must be in #[6,+oo)#

hence

#sqrtx=x-6#

#(sqrtx)^2=(x-6)^2#

#x=(x-6)^2#

#x=x^2-12x+36#

#x^2-12x-x+36=0#

#x^2-13x+36=0#

#(x-4)*(x-9)=0#

#x_1=4# and #x_2=9#

The only acceptable solution is #x=9#