Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 1$, $|\vec{b}| = 2$ and the angle formed by $\vec{a}$ and $\vec{b}$ is $60^\circ$. Set $\vec{u} = x\vec{a} + \vec{b}$ and $\vec{v} = x\vec{a} - \vec{b}$ for a real number $x$. When $x > 1$, we are to find the value of $x$ such that the angle formed by $\vec{u}$ and $\vec{v}$ is $30^\circ$. In the following, $\vec{u} \cdot \vec{v}$ denotes the inner product of $\vec{u}$ and $\vec{v}$, and $\vec{a} \cdot \vec{b}$ denotes the inner product of $\vec{a}$ and $\vec{b}$.
First of all, since the angle formed by $\vec{u}$ and $\vec{v}$ is $30^\circ$, we obtain $$(\vec{u} \cdot \vec{v})^2 = \frac{\mathbf{A}}{\mathbf{B}}|\vec{u}|^2|\vec{v}|^2.$$
When we express this equation in terms of $x$, noting $\vec{a} \cdot \vec{b} = \mathbf{C}$, we have $$x^4 - \mathbf{DE}x^2 + \mathbf{FG} = 0.$$
By transforming this, we also have $$\left(x^2 - \mathbf{H}\right)^2 = (\mathbf{I}x)^2.$$
When this is solved for $x$, we obtain $$x = \mathbf{J} + \sqrt{\mathbf{KL}},$$ noting $x > 1$.