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

The correct option is B $1+\sqrt{3}$

Using cosine rule for $\angle C$

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

$\frac{\sqrt{3}}{2}=\frac{(x^2+x+1{)}^2+(x^2-1{)}^2-(2x+1{)}^2}{2(x^2+x+1)(x^2-1)}=\frac{(2x^4+2x^3-3x^2-2x+1)}{2(x^2+x+1)(x^2-1)}=\frac{(2x^2+2x-1)(x^2-1)}{2(x^2+x+1)(x^2-1)}$

$\Rightarrow \sqrt{3}=\frac{2x^2+2x-1}{x^2+x+1}$

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

$\Rightarrow x=\frac{(2-\sqrt{3})\pm \sqrt{3}}{2(\sqrt{3}-2)}$

$\Rightarrow x=-(2+\sqrt{3})$ , $1+\sqrt{3}\Rightarrow x=1+\sqrt{3}$ as $(x>0)$.