For what values of x is the following expression a real number? \( \frac{7}{\sqrt{16-2 x}} \)

To determine for which values of \( x \) the expression \[ \frac{7}{\sqrt{16 - 2x}} \] is a real number, we need to ensure that the denominator is both **real** and **non-zero**. ### Step-by-Step Analysis: 1. **Denominator Must Be Real:** - The expression inside the square root, \( 16 - 2x \), must be non-negative because the square root of a negative number is not a real number. \[ 16 - 2x \geq 0 \] 2. **Solve the Inequality:** \[ 16 - 2x > 0 \quad \text{(strict inequality because the denominator cannot be zero)} \] \[ -2x > -16 \] \[ x < 8 \] 3. **Conclusion:** - For all real numbers \( x \) that are less than 8, the expression under the square root is positive, ensuring the denominator is real and non-zero. ### Final Answer: All real numbers x with x below eight. In other words, the expression is real for all x < 8