## Precalculus (6th Edition)

The solution is $x=9$
$\sqrt{x}-\sqrt{x-5}=1$ Take $\sqrt{x-5}$ to the right side: $\sqrt{x}=1+\sqrt{x-5}$ Square both sides: $(\sqrt{x})^{2}=(1+\sqrt{x-5})^{2}$ $x=1+2\sqrt{x-5}+x-5$ Leave the square root alone on the right side: $x-1-x+5=2\sqrt{x-5}$ $4=2\sqrt{x-5}$ Take $2$ to divide the left side: $\dfrac{4}{2}=\sqrt{x-5}$ $2=\sqrt{x-5}$ Square both sides: $2^{2}=(\sqrt{x-5})^{2}$ $4=x-5$ Rearrange: $x-5=4$ Solve for $x$: $x=4+5$ $x=9$ The solution is $x=9$