√(3x - 6)/(1 - x)+ 5 > 0

To solve the inequality \(\sqrt{\frac{3x - 6}{1 - x}} + 5 > 0\), we can start by isolating the square root term. ### Step 1: Isolate the Square Root Subtract 5 from both sides: \[ \sqrt{\frac{3x - 6}{1 - x}} > -5 \] Since the square root is always non-negative, this inequality is always true. Thus, we need to ensure that the expression inside the square root is defined and non-negative. ### Step 2: Set the Inequality Inside the Square Root We need to solve: \[ \frac{3x - 6}{1 - x} \geq 0 \] ### Step 3: Find Critical Points The critical points occur where the numerator and denominator are zero: 1. **Numerator**: \(3x - 6 = 0\) \[ 3x = 6 \implies x = 2 \] 2. **Denominator**: \(1 - x = 0\) \[ x = 1 \] ### Step 4: Test Intervals We will analyze the sign of the expression \(\frac{3x - 6}{1 - x}\) in the intervals determined by the critical points \(x = 1\) and \(x = 2\). - **Intervals**: \((-\infty, 1)\), \((1, 2)\), \((2, \infty)\) 1. **Interval \((-\infty, 1)\)**: Choose \(x = 0\) \[ \frac{3(0) - 6}{1 - 0} = \frac{-6}{1} < 0 \quad \text{(negative)} \] 2. **Interval \((1, 2)\)**: Choose \(x = 1.5\) \[ \frac{3(1.5) - 6}{1 - 1.5} = \frac{4.5 - 6}{1 - 1.5} = \frac{-1.5}{-0.5} = 3 \quad \text{(positive)} \] 3. **Interval \((2, \infty)\)**: Choose \(x = 3\) \[ \frac{3(3) - 6}{1 - 3} = \frac{9 - 6}{1 - 3} = \frac{3}{-2} < 0 \quad \text{(negative)} \] ### Step 5: Determine the Valid Intervals From the tests, we find that \(\frac{3x - 6}{1 - x} \geq 0\) in the interval \((1, 2]\). ### Step 6: Conclusion Thus, the solution to the inequality \(\sqrt{\frac{3x - 6}{1 - x}} + 5 > 0\) is: \[ x \in (1, 2] \] This means \(x\) can take any value greater than 1 and less than or equal to 2.