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x\left(2x+3\right)<0
Factor out x.
x+\frac{3}{2}>0 x<0
For the product to be negative, x+\frac{3}{2} and x have to be of the opposite signs. Consider the case when x+\frac{3}{2} is positive and x is negative.
x\in \left(-\frac{3}{2},0\right)
The solution satisfying both inequalities is x\in \left(-\frac{3}{2},0\right).
x>0 x+\frac{3}{2}<0
Consider the case when x is positive and x+\frac{3}{2} is negative.
x\in \emptyset
This is false for any x.
x\in \left(-\frac{3}{2},0\right)
The final solution is the union of the obtained solutions.