Answer
$x\in \{1, 5\}$
Work Step by Step
$\sqrt{2x-1}-\sqrt{x-1}=1$
$(\sqrt{2x-1})^2=(1+\sqrt{x-1})^2$
$2x-1=2\sqrt{x-1}+x$
$x-1=2\sqrt{x-1}$
$x^2-2x+1=4(x-1)$
$x^2-6x+5=0$
$x^2-x-5x+5=0$
$x(x-1)-5(x-1)=0$
$(x-5)(x-1)=0$
$x\in \{1, 5\}$
Check the solutions by substituting each of them in the given equation.
