How do you solve #sqrt(x-2)=x-2#?

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Answer 1 · 2017-07-19T11:54:59

#x = 2# or #x=3#

#sqrt(x-2) = x-2#

If we let #alpha = x-2#, then #sqrtalpha = alpha#. We know that this can only be the case if #alpha = 0# or #alpha =1#. Then #x = 0+2# or #x=1+2#.

Answer 2 · 2017-07-19T11:57:48

The solutions are #S={2,3}#

Squaring the LHS and the RHS

#sqrt(x-2)=x-2#

#(sqrt(x-2))^2=(x-2)^2#

#x^2-4x+4=color(red)(x-2)#

#x^2-4xcolor(red)(-x)+4color(red)(+2)=0#

#x^2-5x+6=0#

Factorising

#(x-2)(x-3)=0#

Therefore,

#(x-2)=0#, #=>#, #x=2#

#(x-3)=0#, #=>#, #x=3#

Verification

#sqrt(2-2)=2-2#

#sqrt(3-2)=3-2#