The middle point of the interval in which `x^(2)+2 (sqrt(x))^(2)-3 le 0` is

Options

A. `1//2`
B. 1
C. 0
D. `-1//2`

Answer

To solve the inequality \( x^2 + 2(\sqrt{x})^2 - 3 \leq 0 \) and find the midpoint of the interval in which this inequality holds, we can follow these steps:

### Step 1: Rewrite the Inequality
Start by rewriting the inequality:
\[ x^2 + 2(\sqrt{x})^2 - 3 \leq 0 \]
Since \((\sqrt{x})^2 = x\), we can simplify this to:
\[ x^2 + 2x - 3 \leq 0 \]

### Step 2: Factor the Quadratic Expression
Next, we need to factor the quadratic expression \(x^2 + 2x - 3\):
\[ x^2 + 2x - 3 = (x + 3)(x - 1) \]
Thus, the inequality becomes:
\[ (x + 3)(x - 1) \leq 0 \]

### Step 3: Find the Roots
To find the intervals where the product is less than or equal to zero, we first find the roots of the equation:
\[ (x + 3) = 0 \quad \Rightarrow \quad x = -3 \]
\[ (x - 1) = 0 \quad \Rightarrow \quad x = 1 \]
The roots are \(x = -3\) and \(x = 1\).

### Step 4: Determine the Intervals
Now we can test the intervals determined by these roots:
1. \( (-\infty, -3) \)
2. \( (-3, 1) \)
3. \( (1, \infty) \)

### Step 5: Test the Intervals
- For \(x < -3\) (e.g., \(x = -4\)):
\((x + 3)(x - 1) = (-)(-) = +\) (not satisfying)

- For \(-3 < x < 1\) (e.g., \(x = 0\)):
\((x + 3)(x - 1) = (+)(-) = -\) (satisfying)

- For \(x > 1\) (e.g., \(x = 2\)):
\((x + 3)(x - 1) = (+)(+) = +\) (not satisfying)

### Step 6: Include the Roots
Since the inequality is less than or equal to zero, we include the roots:
The solution to the inequality is:
\[ [-3, 1] \]

### Step 7: Find the Midpoint
To find the midpoint of the interval \([-3, 1]\), we use the formula for the midpoint:
\[ \text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} = \frac{-3 + 1}{2} = \frac{-2}{2} = -1 \]

### Conclusion
The midpoint of the interval in which \(x^2 + 2(\sqrt{x})^2 - 3 \leq 0\) is:
\[ \boxed{-1} \]