Chapter 1.7, Problem 37E

Is this reasoning for finding the solutions of $\sqrt{x+3}=3-x$ correct? $\left(1\right)\sqrt{x+3}=3-x$ is given; $\left(2\right)x+3=x^2-6x+9$ , obtained by squaring both sides of $\left(1\right);\left(3\right)0=x^2-7x+6$ , obtained by subtracting $x+3$ from both sides of $\left(2\right);\left(4\right)0=\left(x-1\right)\left(x-6\right)$ , obtained by factoring the right-hand side of $\left(3\right);\left(5\right)x=1$ or $x=6$ , which follows from (4) because $ab=0$ implies that $a=0$ or $b=0$ .