11.1 For what values of the parameter $a$ does the equation $x^{2}-6 a x-2+2 a+9 a^{2}=0$ have at least one negative root?

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Answer. $a3 a$. For $a0$. Solving it, we get $(-1-\sqrt{19}) / 9<a<(-1+\sqrt{19}) / 9$. Combining this with the solution for $a<0$, we obtain the answer. Another solution method (more visual) is based on considering two cases depending on the sign of the abscissa $x=3 a$ of the vertex of the parabola $y=x^{2}-6 a x-2+2 a+9 a^{2}$. Since the branches of the parabola are directed upwards, in the first case (for $a<0$) the condition of the problem is equivalent to the ordinate of the vertex being negative, i.e., $y(3 a)<0$. In the second case (for $a \geq 0$) the condition of the problem is equivalent to the inequality $y(0)<0$, and we arrive at the same inequalities as above.