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: (A), (B)

$-(2+\sqrt{3})$
$1+\sqrt{3}$

We have,

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

$\cos \left(\frac{\pi }{6}\right)=\frac{(x^2+x+1{)}^2+(x^2-1{)}^2-(2x+1{)}^2}{2(x^2+x+1)(x^2-1)}$

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

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

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

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

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

On solving we get

$x^2+x-(3\sqrt{3}+5)=0$

$x=\sqrt{3}+1,-(2+\sqrt{3})$