Suppose \( |\vec{a}| = k \) and \( |\vec{b}| = 2k \). Also, the given equation is: \[ k^2 + x(2k^2)\cos 120^\circ - x(2k)^2 - k(2k)\cos 120^\circ = 0 \] We know that \( \cos 120^\circ = -\frac{1}{2} \), so substituting this in: \[ k^2 + x(2k^2)\left(-\frac{1}{2}\right) - x(2k)^2 - k(2k)\left(-\frac{1}{2}\right) = 0 \] Simplifying: \[ k^2 - xk^2 - 4xk^2 + k^2 = 0 \] \[ 2k^2 - 5xk^2 = 0 \] Dividing through by \( k^2 \) (assuming \( k \neq 0 \)): \[ 2 - 5x = 0 \] Solving for \( x \): \[ x = \frac{2}{5} \]