## College Algebra (11th Edition)

$x=9$
We are given: $\sqrt{x}-\sqrt{x-5}=1$ $\sqrt{x}=1+\sqrt{x-5}$ We square both sides: $(\sqrt{x})^{2}=(1+\sqrt{x-5})^{2}$ $(\sqrt{x})^{2}=(1+\sqrt{x-5})(1+\sqrt{x-5})$ We distribute: $x=1+2\sqrt{x-5}+(x-5)$ And combine like terms: $0=-4+2\sqrt{x-5}$ $4=2\sqrt{x-5}$ $2=\sqrt{x-5}$ We square both sides again: $2^{2}=(\sqrt{x-5})^{2}$ $4=x-5$ And solve for $x$: $x=9$