solve `sqrt(x+2) ge x `

Explanation

To solve the inequality \( \sqrt{x + 2} \geq x \), we will follow these steps:

### Step 1: Square both sides
We start with the inequality:
\[ \sqrt{x + 2} \geq x \]
Squaring both sides (noting that squaring is valid since both sides are non-negative):
\[ x + 2 \geq x^2 \]

### Step 2: Rearrange the inequality
Rearranging the inequality gives:
\[ 0 \geq x^2 - x - 2 \]
or
\[ x^2 - x - 2 \leq 0 \]

### Step 3: Factor the quadratic
Next, we factor the quadratic expression:
\[ x^2 - x - 2 = (x - 2)(x + 1) \]
Thus, we rewrite our inequality as:
\[ (x - 2)(x + 1) \leq 0 \]

### Step 4: Find the roots
The roots of the equation \( (x - 2)(x + 1) = 0 \) are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 5: Test intervals
We will test the sign of \( (x - 2)(x + 1) \) in the intervals determined by the roots:
1. \( (-\infty, -1) \)
2. \( (-1, 2) \)
3. \( (2, \infty) \)

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 2)(-2 + 1) = (-4)(-1) = 4 \quad (\text{positive}) \]

- For \( -1 < x < 2 \) (e.g., \( x = 0 \)):
\[ (0 - 2)(0 + 1) = (-2)(1) = -2 \quad (\text{negative}) \]

- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ (3 - 2)(3 + 1) = (1)(4) = 4 \quad (\text{positive}) \]

### Step 6: Determine the solution set
The expression \( (x - 2)(x + 1) \leq 0 \) is satisfied in the interval:
\[ [-1, 2] \]

### Step 7: Consider the domain of the original inequality
Since we have a square root in the original inequality, we must ensure that \( x + 2 \geq 0 \):
\[ x + 2 \geq 0 \implies x \geq -2 \]

### Step 8: Combine the intervals
The solution must satisfy both conditions:
1. \( x \in [-1, 2] \)
2. \( x \geq -2 \)

Thus, the solution set is:
\[ [-1, 2] \]

### Final Answer
The final solution to the inequality \( \sqrt{x + 2} \geq x \) is:
\[ x \in [-1, 2] \]

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