- #1

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- Homework Statement
- Solve : ##\sqrt{x+2} \ge x##

- Relevant Equations
- For ##\sqrt{f(x)}##, we must have both (1) ##\sqrt{f(x)} \ge 0## and (2) ##f(x)\ge 0##

**Let me copy and paste the problem as it appears in the text on the right.
Problem statement :**

**Attempt (myself) :**By looking at ##\large{\sqrt{x+2}\ge x}##, from my Relevant Equations above, we have the following :

1. Outcome ##\mathbf{x \ge 0}##, since square roots are always positive.

2. Function inside the square root, ##x+2 \ge 0\Rightarrow \mathbf{x \ge -2}##.

Armed with these inequalities, I square both sides of it. ##x+2 \ge x^2\Rightarrow x^2 - x - 2 \le 0\Rightarrow (x-2)(x+1)\le 0\Rightarrow \mathbf{-1\le x \le 2}##.

Looking at the three (

**bold faced**) solutions above and merging them, I find that the answer must be : ##\boxed{\mathbf{0 \le x \le 2\Rightarrow x \in [0,2]}}##.

But the book has a different answer.

**Attempt (Text Solution) :**I copy and paste below the solution to the problem as it appears in the text.

**Doubt :**Of course is the text correct. For instance how can ##x## lie in the interval ##[-2,2]##? That would imply that ##x## can be equal to -1, which would make the square root of a quantity negative.

**A hint or suggestion would be welcome as to where have I gone wrong.**