Question

Let $A B C$ be a triangle such that $\angle A C B=\frac{\pi}{6}$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value(s) of $x$ for which $a=x^2+x+1, b=x^2-1$ and $c=2 x+1$ is/are

• A. $-(2+\sqrt{3})$
• B. $1+\sqrt{3}$
• C. $2+\sqrt{3}$
• D. $4 \sqrt{3}$

Answer 1

Answer: (B)

Using $\cos C=\frac{a^2+b^2-c^2}{2 a b}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\left(x^2+x+1\right)^2+\left(x^2-1\right)^2-(2 x+1)^2}{2\left(x^2+x+1\right)\left(x^2-1\right)}$
$\Rightarrow (x+2)(x+1)(x-1) x+\left(x^2-1\right)^2 $
$=\sqrt{3}\left(x^2+x+1\right)\left(x^2-1\right)$
$\Rightarrow x^2+2 x+\left(x^2-1\right)=\sqrt{3}\left(x^2+x+1\right)$
$\Rightarrow (2-\sqrt{3}) x^2+(2-\sqrt{3}) x-(\sqrt{3}+1)=0$
$ \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.