3. Solve for \( x \). 3.1. \( \sqrt{x-2}=4-x \) 3.2. \( A=\frac{\sqrt{x+4}}{x-2} \) : For which values of \( x \) will A be: 3.2.1 Non real? (2) 3.2.2. Undefined? (2) 3-2-3. zero ? (2.)

To solve \( \sqrt{x-2}=4-x \), we start by squaring both sides: \( x - 2 = (4 - x)^2 \) Expanding the right-hand side gives us: \( x - 2 = 16 - 8x + x^2 \) Rearranging gives us: \( x^2 - 9x + 18 = 0 \) Factoring or using the quadratic formula leads us to the solutions \( x = 6 \) and \( x = 3 \). However, we need to check these solutions in the original equation. Plugging in \( x = 6 \) works, while \( x = 3 \) fails to satisfy the equation, leaving us with \( x = 6 \) as the only valid solution. Next, let's examine \( A=\frac{\sqrt{x+4}}{x-2} \): 1. **For \( A \) to be non-real**, the expression under the square root must be negative, suggesting \( x+4 < 0 \) or \( x < -4 \). Therefore, any \( x < -4 \) will make \( A \) non-real since it causes the square root to behave oddly! 2. **For \( A \) to be undefined**, the denominator must equal zero. This occurs when \( x-2=0 \rightarrow x=2 \). So, when \( x = 2 \), the expression is undefined, running away from the land of valid numbers! 3. **For \( A \) to be zero**, the numerator must equal zero while the denominator is not. This happens when \( \sqrt{x+4}=0 \rightarrow x+4=0 \rightarrow x=-4 \). Thus, the value of \( x = -4 \) will make \( A \) zero while keeping it defined! And voila! You’ve navigated through equations like a mathematical ninja!