Question

Let $ABC$ be a triangle such that $\angle ACB=\frac{\pi }{6}$ 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 $c=2x+1$ is/are

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

Answer 1

Answer: (B)

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

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{{\left(x^2+x+1\right)}^2+{\left(x^2-1\right)}^2-(2x+1{)}^2}{2\left(x^2+x+1\right)\left(x^2-1\right)}$
$\Rightarrow (x+2)(x+1)(x-1)x+{\left(x^2-1\right)}^2$
$=\sqrt{3}\left(x^2+x+1\right)\left(x^2-1\right)$
$\Rightarrow x^2+2x+\left(x^2-1\right)=\sqrt{3}\left(x^2+x+1\right)$
$\Rightarrow (2-\sqrt{3})x^2+(2-\sqrt{3})x-(\sqrt{3}+1)=0$
$\Rightarrow x=-(2+\sqrt{3})$
and $x=1+\sqrt{3}$
But, $x=-(2+\sqrt{3})\Rightarrow c$ is negative.
$\therefore x=1+\sqrt{3}$ is the only solution.