Let ABC be a triangle such that ∠ACB = π/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 = x² + x + 1, b = x² − 1 and c = 2x + 1 is (are)
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Using cosine rule for ∠C,
√3/2 = ((x² + x + 1)² + (x² - 1)² - (2x + 1)²)/(2*(x² + x + 1)*(x² - 1))
⇒ √3 = (2x² + 2x - 1)/(x² + x + 1)
⇒ (√3 - 2)x² + (√3 - 2)x + (√3 + 1) = 0
⇒ x = ((2 - √3) ± √3 ) / (2 * (√3 - 2))
⇒ x = -(2 + √3), 1 + √3
⇒ x = 1 + √3 as (x > 0)
Hope this helps.