Solution assignment 03 Root equations

Solve:

$\sqrt{x-1}=x$

The following conditions have to be met:

$x\geq1$ and $x\geq0$

thus:

$x\geq1$

Squaring both sides yields:

$x-1=x^2$

$x^2-x+1=0$

This quadratic equation has a discriminant less than $0$ :

$D=1-4=-3<0$

and thus the equation has no solutions. This also holds for the original equation.