Solve For \( x \) 1. \( 2 \sqrt{x+2}=x-1 \)

To solve the equation \( 2 \sqrt{x+2} = x - 1 \), first isolate the square root by dividing both sides by 2: \[ \sqrt{x+2} = \frac{x-1}{2} \] Next, square both sides to eliminate the square root: \[ x + 2 = \left(\frac{x-1}{2}\right)^2 \] Expanding the right side gives: \[ x + 2 = \frac{(x-1)^2}{4} \] Now, multiply both sides by 4 to eliminate the fraction: \[ 4(x + 2) = (x-1)^2 \] This expands to: \[ 4x + 8 = x^2 - 2x + 1 \] Rearranging the equation brings all terms to one side: \[ 0 = x^2 - 2x + 1 - 4x - 8 \] Combining like terms results in: \[ 0 = x^2 - 6x - 7 \] Now, factor the quadratic equation: \[ 0 = (x - 7)(x + 1) \] Setting each factor to zero gives the solutions: \[ x - 7 = 0 \quad \Rightarrow \quad x = 7 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] Finally, we need to check both solutions in the original equation. For \(x = 7\): \[ 2 \sqrt{7 + 2} = 2 \sqrt{9} = 2 \cdot 3 = 6 \] \[ 7 - 1 = 6 \quad \Rightarrow \quad \text{Valid solution!} \] For \(x = -1\): \[ 2 \sqrt{-1 + 2} = 2 \sqrt{1} = 2 \] \[ -1 - 1 = -2 \quad \Rightarrow \quad \text{Invalid solution!} \] Thus, the only solution is: \[ \boxed{7} \]