If 9 - x^2 ≥ 0, which of the following describes...

$$If\ 9-x^2\ge0,$$
Which of the following describes all possible values of x?

$$A.\ 3\ge x\ge-3$$
$$B.\ x\ge3\ or\ x\le-3$$
$$C.\ 3\ge x\ge0$$
$$D.\ -3\le x$$
$$E.\ 3\le x$$

The OA is A.

I don't have clear this PS question but I think that this is a simple question that I can solve the following way,
$$9-x^2\ can\ be\ write\ like\ \left(3-x\right)^2,\ right?$$
$$Then,\ \left(3-x\right)^2\ge0\ is\ equivalent\ to\ \left(3-x\right)\left(3+x\right)\ge0$$
Finally, this expression will be true for,
$$x\ge-3\ and\ x\le3$$
Or
$$-3\le x\le3$$
I appreciate if any expert explains it to me. Thank you so much.

Hi AAPL,
Let's take a look at your question.

First of all let me clarify that $$9-x^2\ can\ not\ be\ written\ like\ \left(3-x\right)^2.$$
$$We\ will\ write\ 9-x^2\ as\ difference\ of\ squares\ like:$$
$$\left(3\right)^2-\left(x\right)^2\ge0 ... (i)$$
$$Using\ difference\ of\ square\ formula\ a^2-b^2=\left(a+b\right)\left(a-b\right),\ inequality\ \left(i\right)\ can\ be\ written\ as:$$
$$\left(3+x\right)\left(3-x\right)\ge0$$
$$\left(3+x\right)\ge0\ and\ \left(3-x\right)\ge0$$
$$x\ge-3\ and\ -x\ge-3$$
$$x\ge-3\ and\ x\le3$$
This expression includes all values between -3 and 3 inclusive.
Remember that this can be represented in two more ways using inequality sign as:
$$-3\le x\le3\ or\ 3\ge x\ge-3$$

Therefore, Option A is correct.

Hope it helps.
I am available if you'd like any follow up.