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)

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Aicharya · Accepted Answer

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

$ \begin{gathered} \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 \end{gathered} $ $\Rightarrow x=-(2+\sqrt{3})$ and $x=1+\sqrt{3}$ But, $x=-(2+\sqrt{3}) \Rightarrow c$ is negative. $\therefore \quad x=1+\sqrt{3}$ is the only solution. Hence, (b) is the correct option

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…

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