Solve for x
x\in \left(0,\frac{1}{2}\right)
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2x^{2}-x<0
Subtract x from both sides.
x\left(2x-1\right)<0
Factor out x.
x>0 x-\frac{1}{2}<0
For the product to be negative, x and x-\frac{1}{2} have to be of the opposite signs. Consider the case when x is positive and x-\frac{1}{2} is negative.
x\in \left(0,\frac{1}{2}\right)
The solution satisfying both inequalities is x\in \left(0,\frac{1}{2}\right).
x-\frac{1}{2}>0 x<0
Consider the case when x-\frac{1}{2} is positive and x is negative.
x\in \emptyset
This is false for any x.
x\in \left(0,\frac{1}{2}\right)
The final solution is the union of the obtained solutions.
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