Solve for x
x\in \left(3\sqrt{2},2\sqrt{5}\right)
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x-2\sqrt{5}>0 x-3\sqrt{2}<0
For the product to be negative, x-2\sqrt{5} and x-3\sqrt{2} have to be of the opposite signs. Consider the case when x-2\sqrt{5} is positive and x-3\sqrt{2} is negative.
x\in \emptyset
This is false for any x.
x-3\sqrt{2}>0 x-2\sqrt{5}<0
Consider the case when x-3\sqrt{2} is positive and x-2\sqrt{5} is negative.
x\in \left(3\sqrt{2},2\sqrt{5}\right)
The solution satisfying both inequalities is x\in \left(3\sqrt{2},2\sqrt{5}\right).
x\in \left(3\sqrt{2},2\sqrt{5}\right)
The final solution is the union of the obtained solutions.
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