If \( \alpha \) is a root of the equation \( 2 x(2 x+1)=1 \), find the other \( r \)

First, let's solve the equation \( 2x(2x + 1) = 1 \). We can expand it to get: \[ 4x^2 + 2x - 1 = 0 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 2 \), and \( c = -1 \). Calculating the discriminant: \[ b^2 - 4ac = 2^2 - 4 \cdot 4 \cdot (-1) = 4 + 16 = 20 \] Now substituting into the quadratic formula: \[ x = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4} \] Thus, the roots are: \[ \alpha = \frac{-1 + \sqrt{5}}{4} \quad \text{and} \quad r = \frac{-1 - \sqrt{5}}{4} \] The other root \( r \) is: \[ \boxed{\frac{-1 - \sqrt{5}}{4}} \]