I think it cannot be solved the usual ways.
The x will stretch as it wishes but there are limits to its stretch.
The minimum stretch, or squeeze, would be when x tends to be zero.
If x = 0, there would be no isosceles triangles left, and there would be no separate, the 3rd, triangle. Hence x cannot be zero.
So that the 3rd triangle would exist, the x should tend to become [sqrt(17) -sqrt(11)]
The maximum stretch would be when x tends to be 2sqrt(11).
Then the isosceles triangle with two sqrt(17) sides would be visible, while the isosceles triangle with two sqrt(11) sides would tend to be flat or a straight line only.
The 3rd triangle would exist because any one side would be less than the sum of the other two sides.
[sqsrt(17) -sqrt(11)] < x < 2sqrt(11) -------------answer.