Solve absolute value inequalities depending on the parameter $a$.
$$|x|(ax+1)<2$$ In the first case where we have $x>0,\;a>0$ we get: $$ax^2+x-2<0$$ I get that $x$ is in the interval $\left(0,-\frac{1+\sqrt{1+8a}}{2a}\right)$?
2026-04-03 19:48:35.1775245715
What are the solutions to the inequality $|x|(ax+1)<2$?
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you have to solve $$x>0$$ and $$ax^2+x-2<0$$ or $$x<0$$ and $$-ax^2-x-2<0$$