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Let A B C be a triangle such that /A C B...

Let `A B C` be a triangle such that `/_A C B=pi/6` and let `a` , `b` and `c` denote the lengths of the side 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) `-(2+sqrt(3))` (b) `1+sqrt(3)` (c)`2+sqrt(3)` (d) `4sqrt(3)`

A

`-(2 + sqrt3)`

B

`1+ sqrt3`

C

`2+sqrt3`

D

`4 sqrt3`

Text Solution

Verified by Experts

The correct Answer is:
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)`
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