## College Algebra (6th Edition)

$x=2,6$
Separate the square roots on two sides. Then square both sides and combine like terms. Put all the terms on one side and factor by trial and error. Then solve for $x$ with the values found. $\sqrt {2x-3}-\sqrt {x-2}=1$ $\sqrt {2x-3}=1+\sqrt {x-2}$ $(\sqrt {2x-3})^2=(1+\sqrt {x-2})^2$ $2x-3=1+2\sqrt {x-2}+x-2$ $x-2=2\sqrt {x-2}$ $(x-2)^2=(2\sqrt {x-2})^2$ $x^2-4x+4=4(x-2)$ $x^2-8x+12=0$ $(x-2)(x-6)=0$ $x=2,6$