If $\tan \theta$ and $\cot \theta$ are the roots of the equation $x ^ { 2 } + 2 x + 1 = 0$ then the least value of $x ^ { 2 } + \tan \theta x + \cot \theta = 0$ is

Solving x^2 + 2x + 1 = 0 gives a double root of -1, so tan theta = -1 and cot theta = -1. Substituting these into x^2 + (tan theta)x + cot theta = 0 yields x^2 - x - 1 = 0. The least value of this quadratic is found using the vertex formula, giving -(-1)^2 - 4(1)(-1) all over 4(1), which equals -5/4. The result is -5/4.

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