Mathematics - Properties of Triangles Question with Solution | TestHub

Using, $\cos C=\frac{a^2+b^2-c^2}{2ab}$
$$\begin{array}{c}\Rightarrow \frac{\sqrt{3}}{2}=\frac{{\left(x^2+x+1\right)}^2+{\left(x^2-1\right)}^2-(2x+1{)}^2}{2\left(x^2+x+1\right)\left(x^2-1\right)}\\ \Rightarrow (x+2)(x+1)(x-1)x+{\left(x^2-1\right)}^2\\ =\sqrt{3}\left(x^2+x+1\right)\left(x^2-1\right)\\ \Rightarrow x^2+2x+\left(x^2-1\right)=\sqrt{3}\left(x^2+x+1\right)\\ \Rightarrow (2-\sqrt{3})x^2+(2-\sqrt{3})x-(\sqrt{3}+1)=0\end{array}$$ $\Rightarrow x=-(2+\sqrt{3})$ and $x=1+\sqrt{3}$ But, $x=-(2+\sqrt{3})\Rightarrow c$ is negative. $\therefore x=1+\sqrt{3}$ is the only solution. Hence, (b) is the correct option.