18) Solve \( x+6=\sqrt{2 x+15} \)

Let’s solve the equation \( x+6=\sqrt{2x+15} \). First, we’ll square both sides to eliminate the square root: \[ (x + 6)^2 = 2x + 15 \] Expanding the left side gives us: \[ x^2 + 12x + 36 = 2x + 15 \] Now, let's rearrange all terms to one side: \[ x^2 + 12x + 36 - 2x - 15 = 0 \] This simplifies to: \[ x^2 + 10x + 21 = 0 \] Next, we can factor this quadratic equation: \[ (x + 3)(x + 7) = 0 \] Setting each factor to zero gives us the solutions: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] \[ x + 7 = 0 \quad \Rightarrow \quad x = -7 \] Now, it's important to check both solutions in the original equation to ensure they are valid. 1. For \( x = -3 \): \[ -3 + 6 = \sqrt{2(-3) + 15} \] \[ 3 = \sqrt{-6 + 15} \] \[ 3 = \sqrt{9} \quad \Rightarrow \quad 3 = 3 \quad \text{(valid)} \] 2. For \( x = -7 \): \[ -7 + 6 = \sqrt{2(-7) + 15} \] \[ -1 = \sqrt{-14 + 15} \] \[ -1 = \sqrt{1} \quad \Rightarrow \quad -1 \neq 1 \quad \text{(not valid)} \] Thus, the only solution is: \[ \boxed{-3} \]