Solve for x
x\in \begin{bmatrix}-\frac{1}{5},0\end{bmatrix}
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x\left(5x+1\right)\leq 0
Factor out x.
x+\frac{1}{5}\geq 0 x\leq 0
For the product to be ≤0, one of the values x+\frac{1}{5} and x has to be ≥0 and the other has to be ≤0. Consider the case when x+\frac{1}{5}\geq 0 and x\leq 0.
x\in \begin{bmatrix}-\frac{1}{5},0\end{bmatrix}
The solution satisfying both inequalities is x\in \left[-\frac{1}{5},0\right].
x\geq 0 x+\frac{1}{5}\leq 0
Consider the case when x+\frac{1}{5}\leq 0 and x\geq 0.
x\in \emptyset
This is false for any x.
x\in \begin{bmatrix}-\frac{1}{5},0\end{bmatrix}
The final solution is the union of the obtained solutions.
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