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 $2$ .

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