In the absence of a diagram, we can use the cosine rule to figure out the range for x:
(2x+31)^2=(2x+1)^2+(x+16)^2-2(2x+1)(x+16)cosA where A is an angle.
(2x+31)^2-(2x+1)^2=(x+16)(x+16-(4x+2)cosA);
(2x+31-2x-1)(2x+31+2x+1)=(x+16)(x+16-(4x+2)cosA);
30(4x+32)=(x+16)(x+16-(4x+2)cosA).
cosA must lie between -1 and 1: -1<cosA<1; or -1<(x^2-88x-704)/(2(x+16)(2x+1))<1;
-(4x^2+66x+32)<x^2-88x-704<4x^2+66x+32 (this cross-multiplication is only valid if the denominator is positive; otherwise the inequality has to be reversed; assume for now that it is positive);
5x^2-22x-672>0, (5x+48)(x-14)>0.
So, x>14 or x<-48/5; note that x>14 makes 2(x+16)(2x+1) positive, but x<-48/5 makes it negative, so the inequality is reversed and x>-48/5.
Alternatively:
3x^2+154x+736>0, (3x+16)(x+46)>0. This has solutions x<-46 and x>-16/3; the first makes 2(x+16)(2x+1) positive and the second makes it negative, so the inequality must be reversed, and x<-16/3.
We can see that (x^2-88x-704)/(2(x+16)(2x+1)) lies between the required limits for cosA when -48/5<x<-16/3 and x>14, and x<-46, which can be rejected because it would result in all the sides of the triangle being negative.
For the remaining ranges we have the following values for (x+16, 2x+1, 2x+31) on (-48/5,-16/3,14):
(6.4,-18.2,11.8); (10.67,-9.67,20.33); (30,29,59). The first two of these can be rejected because the sides of the triangle must be positive, leaving x>14 as the only solution.