In a triangle ABC, a = 4, b = 3, ∠BAC = 60°, then the equation for which c is the root, is
\[ \text{Given: } a = 4, \quad b = 3, \quad \angle BAC = 60^\circ \] Using the cosine formula, we get: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \] \[ \Rightarrow \cos 60 = \frac{3^2 + c^2 - 4^2}{2 \times 3 \times c} \] \[ \Rightarrow \frac{1}{2} \times 6c = 9 + c^2 - 16 \] \[ \Rightarrow c^2 - 3c - 7 = 0 \]
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