Let ABC be a triangle such that `angleACB=pi/6` and let a , b and c denote the lengths of the side opposite to A ,B and C respectively. The value of x for which `a=x^(2)+x+1,b=x^(2)-1` and `c=2x+1` is

Options

A. `-(2+sqrt3)`
B. `1+sqrt3`
C. `2+sqrt3`
D. `4sqrt3`

Explanation

Using Cosine rule, we have
`cosC=(a^(2)+b^(2)-c^(2))/(2ab)`
`rArrcospi/6=((x^(2)+x+1)^(2)+(x^(2)-1)^(2)-(2x+1)^(2))/(2(x^(2)+x+1)(x^(2)-1))`
`rArrsqrt3=(2x^(2)+2x-1)/(x^(2)+x+1)`
`rArrx^(2)(2-sqrt3)+x(2-sqrt3)-(sqrt3+1)=0`
`x=(-(2-sqrt3)pmsqrt((2-sqrt)^(2)+4(2-sqrt3)(sqrt3+1)))/(2(2-sqrt3))`
`rArrx=-(2+sqrt3),(sqrt3+1)rArrx=sqrt3+1`