Discrete Math Proofs Are these steps for finding the solutions of $\sqrt{x + 3} = 3 − x$ correct? 


$\sqrt{x + 3} = 3 − x$ is given;
$x + 3 = x^2 − 6x + 9$, obtained by squaring both sides of (1);
$0=x^2 −7x+6$, obtained by subtracting $x+3$ from both sides of (2);
$0 = (x − 1)(x − 6)$, obtained by factoring the right-hand side of (3);
$x = 1$ or $x = 6$, which follows from (4) because $ab = 0$ implies that $a=0$ or $b=0$.
Yes, all the steps are correct. You only need to check those value of $x$ if it satisfies the original equation or not.

Assume that $ab=0$. We want to show that either $a=0$ or $b=0$. If $a=0$ then we are done. Suppose that $a\neq 0$. Then $a^{-1}\in\Bbb R$. Thus, using field properties of $\Bbb R$ we get
$$b=1b=(a^{-1}a)b=a^{-1}(ab)=a^{-1}0=0.$$
It's only valid to square the equation when both sides are positive.

In your second line you need to have a condition $3-x\geq 0 $ which simplifies to $x\leq 3$.Also you must have your square root defined i.e $x+3\geq 0$ or $x\geq -3$.

Now that we've done that we can see that $x=6$ is not a solution to the equation,this can be easily verified by plugging $x=6$ or differently $\sqrt{6+3}\neq 3-6$