## Precalculus (6th Edition)

Square both sides to obtain: $(\sqrt{x+2}+5)^2=(\sqrt{x+15})^2 \\(\sqrt{x+2}^2+2(\sqrt{x+2})(5)+5^2=x+15 \\x+2+10\sqrt{x+2}+25=x+15 \\x+10\sqrt{x+2}+27=x+15$ Isolate the terms with $x$ on the left side to obtain: $x+10\sqrt{x+2}-x=15-27 \\10\sqrt{x+2}=-12$ Divide 10 to both sides: $\dfrac{10\sqrt{x+2}}{10}=\dfrac{-12}{10} \\\sqrt{x+2}=-\dfrac{6}{5}$ RECALL: The square root of any real number is greater than or equal to zero. Thus, the value of $\sqrt{x+2}$ is greater than or equal to zero. This means that the value of $\sqrt{x+2}$ cannot be negative. Therefore, the given equation has no solution.