I just want to make sure i have this problem right. It will be posted below.

Question

anonymous · Answer

$$\sqrt{x-3}\ge7$$ I got$$3\le x \le 52$$ or is it just $$x \ge 52$$

kva1992 · Answer

the second one

kva1992 · Answer

because you added the 3 to the other side thus removing the 3 and leaving the x

solomonzelman · Answer

it is just  x≥52

solomonzelman · Answer

$\normalsize\color{blue}{ \sqrt{x-3}≥7 }$   square both sides,

$\normalsize\color{blue}{ x-3≥49 }$

$\normalsize\color{blue}{ x≥52 }$

kva1992 · Answer

this would be the only time you would get the first answer you posted $$0 \le x-3  \le 49$$

kva1992 · Answer

hope i was able to help

solomonzelman · Answer

Bree Bree, you just confused it a little with solving similar equations with absolute values, where the x solutions (or the domain) is limited to he smallest AND biggest value that it can equal.
Not just limited one way, and going infinitely to the other, like in a case of a square root .