\( \sqrt {x+\sqrt {x+\sqrt {x+\sqrt {x+\sqrt {x+\sqrt {x.....}}}}}} \)

To solve the expression \[ y = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}}}} \] we can set up an equation based on the recursive nature of the expression. Since the expression inside the square root is the same as \(y\), we can write: \[ y = \sqrt{x + y} \] Next, we will square both sides to eliminate the square root: \[ y^2 = x + y \] Now, we can rearrange this equation to form a standard quadratic equation: \[ y^2 - y - x = 0 \] We can solve this quadratic equation using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = -1\), and \(c = -x\). Plugging in these values, we get: \[ y = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-x)}}{2 \cdot 1} \] This simplifies to: \[ y = \frac{1 \pm \sqrt{1 + 4x}}{2} \] Since \(y\) represents a length (as it is derived from a square root), we only consider the positive root: \[ y = \frac{1 + \sqrt{1 + 4x}}{2} \] Thus, the solution to the original expression is: \[ \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} = \frac{1 + \sqrt{1 + 4x}}{2} \]