How do you solve #sqrt(x - 2) = x# and find any extraneous solutions?

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Answer 1 · 2017-06-27T08:45:30

#x=(1+-isqrt7)/2# and there are no extraneous solutions.

Observe that as we have #sqrt(x-2)#, we cannot have #x<2# and now to solve

#sqrt(x-2)=x# we square it and write it as

#x-2=x^2#

or #x^2-x+2=0#

As discriminant #Delta=b^2-4ac=(-1)^2-4xx1xx2=-7#

Hencewe have complex roots

#x=(-(-1)+-sqrt(-7))/2=(1+-isqrt7)/2#

and as roots are complex there are no extraneous solutions*.

*as none is real root and less than #2#.