Question

Let `A B C` be a triangle such that `/_A C B=pi/6` and let `a , ba n dc` denote the lengths of the side opposite to `A , B ,a n dC` respectively. The value(s) of `x` for which `a=x^2+x+1,b=x^2-1,a n dc=2x+1` is(are) `-(2+sqrt(3))` (b) `1+sqrt(3)` `2+sqrt(3)` (d) `4sqrt(3)`
A. `-(2 + sqrt3)`
B. `1+ sqrt3`
C. `2+sqrt3`
D. `4 sqrt3`

Answer 1

Correct Answer - B
Using cosine rule for `angleC`, we get
`(sqrt3)/(2) = ((x^(2) + x+1)^(2) + (x^(2) -1)^(2) - (2x + 1)^(2))/(2 (x^(2) + x + 1) (x^(2) -1))`
or `sqrt3 = (2x^(2) + 2x -1)/(x^(2) + x + 1)`
or `(sqrt3 -2) x^(2) + (sqrt3 -2) x + (sqrt3 +1) = 0`
or `x = ((2 -sqrt3) +- sqrt3)/(2(sqrt3 -2))`
or `x = -(2 + sqrt3), 1 + sqrt3`
or `x = 1 + sqrt3 " as " (x gt 0)`