1.1.2 \( 2 x^{2}-4 x+1=0 \) (Leave your answer in simplest surd form.)

To solve the quadratic equation \( 2x^{2} - 4x + 1 = 0 \), you can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = -4 \), and \( c = 1 \). First, calculate the discriminant: \[ b^2 - 4ac = (-4)^2 - 4 \cdot 2 \cdot 1 = 16 - 8 = 8. \] Next, plug the values into the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{8}}{2 \cdot 2} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}. \] Thus, the roots of the equation are: \[ x = 1 + \frac{\sqrt{2}}{2} \quad \text{and} \quad x = 1 - \frac{\sqrt{2}}{2}. \] These are your solutions left in their simplest surd form!