\( (x+2)^{\frac{1}{2}}=x \) What is the solution to the given equation?

To solve the equation: \[ \sqrt{x + 2} = x \] **Step 1: Determine the Domain** The square root function requires the argument to be non-negative: \[ x + 2 \geq 0 \implies x \geq -2 \] Additionally, since \(\sqrt{x + 2} = x\) and the square root is always non-negative, \(x\) must also satisfy: \[ x \geq 0 \] So, the domain of the equation is \(x \geq 0\). **Step 2: Square Both Sides** To eliminate the square root, square both sides of the equation: \[ (\sqrt{x + 2})^2 = x^2 \implies x + 2 = x^2 \] **Step 3: Rearrange the Equation** Bring all terms to one side to form a quadratic equation: \[ x^2 - x - 2 = 0 \] **Step 4: Solve the Quadratic Equation** Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the equation \(x^2 - x - 2 = 0\), \(a = 1\), \(b = -1\), and \(c = -2\). Plugging these in: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \] This gives two potential solutions: \[ x = \frac{1 + 3}{2} = 2 \quad \text{and} \quad x = \frac{1 - 3}{2} = -1 \] **Step 5: Verify the Solutions** Recall the domain \(x \geq 0\): - \(x = 2\) is valid. - \(x = -1\) is **not** valid since it doesn't satisfy \(x \geq 0\). **Conclusion** The only real solution to the equation is: \[ \boxed{2} \]