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

Jun 27, 2017

$x = \frac{1 \pm i \sqrt{7}}{2}$ and there are no extraneous solutions.

Explanation:

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 $\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \times 1 \times 2 = - 7$

Hencewe have complex roots

$x = \frac{- \left(- 1\right) \pm \sqrt{- 7}}{2} = \frac{1 \pm i \sqrt{7}}{2}$

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

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