Question

Let ABC be a triangle such that $\angle ACB=\frac{\pi }{6}$ and let $a,b,c$ denote the lengths of the sides opposite to $A,B,C$ 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. $\sqrt{3}+1$
• C. $2+\sqrt{3}$
• D. $4\sqrt{3}$

Answer 1

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

We know, $\cos C=\frac{a^2+b^2-c^2}{2ab}$
$\frac{\sqrt{3}}{2}=\frac{a^2+b^2-c^2}{2ab}$
$=\frac{(x^2+x+1{)}^2+(x^2-1{)}^2-(2x+1{)}^2}{2(x^2-1)(x^2+x+1)}$
$=\frac{x^4+x^2+1+2x^3+2x^2+2x+x^4-2x^2+1-(4x^2+4x+1)}{2(x^3-1)(x+1)}$
$=\frac{2x^4+2x^3+(3x^2-2x^2-4x^2)+(2x-4x)+1)}{2(x^4+x^3-x-1)}$
$=\frac{2x^4+2x^3-3x^2-2x+1}{2(x^4+x^3-x-1)}$
$=\frac{\sqrt{3}}{2}$
$\sqrt{3}(x^4+x^3-x-1)=2x^4+2x^3-3x^2-2x+1$
$(2-\sqrt{3})x^4+(2-\sqrt{3})x^3-3x^2-(2-\sqrt{3})x+1+\sqrt{3}=0$
From the above options, the only value of $x$ satisfying the above equation is
$x=\sqrt{3}+1$.