How to find the restrictions of side length on an obtuse triangle

Question

Question:

In Triangle ABC, the angle ∠ABC is an obtuse angle. The Side AB is 1cm, and the side BC is 3cm. Side AC is (3x+10)/(x+3) cm Find the restriction(s) on x.

I have tried a few different methods, such as using the cosine rule, but they have iether yoelded the wrong answer, or have taken a VERY long time to solve. What would be the simplest, easiest way to solve this problem?

Thank you for helping x

Answer 1

Use 2 conditions:

Since $\angle ABC$ is obtuse, so $$\left({3x+10\over x+3}\right)^2\gt1^2+3^2$$ And triangle inequality$$0\lt{3x+10\over x+3}\lt 1+3$$

Answer 2

Put $|AC|=:s$. Then $s^2=1^2+3^2-6\cos\beta>10$, and the triangle inequality gives $s\leq1+3$, hence $s^2<16$, so that we now face $$10<\left({3x+10\over x+3}\right)^2<16\ .$$ Multiplying through by $(x+3)^2>0$ we get $$10x^2+60x+90<9x^2 +60x +100<16x^2+96 x+144\ ,\tag{1}$$ or $$x^2<10<7x^2+36 x+54\ .$$ The equation $7x^2+36x+44=0$ is solved by $x_1:=-{44\over14}$ and $x_2:=-2$. It follows that the right inequality in $(1)$ is fulfilled for $x<x_1$ and for $x>x_2$ (note that the forbidden $x=-3$ does not belong to the feasible region). Since $\sqrt{10}\doteq3.162>3.142\doteq{44\over14}$ we conclude that the set $S$ of allowed $x$ is the union of two intervals as follows: $$S=\ \left]-\sqrt{10}, -{44\over14}\right[\ \ \cup\ \ \bigl]-2,\sqrt{10}\bigr[\ .$$