Solve `sqrt((x-3)(2-x)) lt sqrt(4x^(2)+12x+11)`.

Answer

To solve the inequality \(\sqrt{(x-3)(2-x)} < \sqrt{4x^2 + 12x + 11}\), we will follow these steps:

### Step 1: Square both sides
We start by squaring both sides of the inequality to eliminate the square roots:
\[ (x-3)(2-x) < 4x^2 + 12x + 11 \]

### Step 2: Expand both sides
Now, we will expand the left side:
\[ (x-3)(2-x) = 2x - x^2 - 6 + 3x = -x^2 + 5x - 6 \]
So, the inequality becomes:
\[ -x^2 + 5x - 6 < 4x^2 + 12x + 11 \]

### Step 3: Move all terms to one side
Next, we will move all terms to one side to set the inequality to zero:
\[ -x^2 - 4x^2 + 5x - 12x - 6 - 11 < 0 \]
This simplifies to:
\[ -5x^2 - 7x - 17 < 0 \]

### Step 4: Multiply by -1
To make the leading coefficient positive, we multiply the entire inequality by -1 (remember to flip the inequality sign):
\[ 5x^2 + 7x + 17 > 0 \]

### Step 5: Analyze the quadratic
Next, we need to determine if the quadratic \(5x^2 + 7x + 17\) is always positive. We can check the discriminant:
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 5 \cdot 17 = 49 - 340 = -291 \]
Since the discriminant is negative, the quadratic has no real roots and opens upwards (as the coefficient of \(x^2\) is positive). Therefore, \(5x^2 + 7x + 17 > 0\) for all \(x\).

### Step 6: Check the conditions for the original inequality
Now we need to ensure that the expressions under the square roots are non-negative:
1. \(x - 3 \geq 0 \Rightarrow x \geq 3\)
2. \(2 - x \geq 0 \Rightarrow x \leq 2\)

However, these two conditions cannot be satisfied simultaneously. Thus, we check the other conditions derived from the product being positive:
- Either both \(x - 3 > 0\) and \(2 - x > 0\) or both \(x - 3 < 0\) and \(2 - x < 0\).

The only valid case is:
- \(x < 3\) and \(x > 2\) which gives us \(2 < x < 3\).

### Final Solution
The solution to the inequality is:
\[ x \in (2, 3) \]