If the equation x^2 + 1 x^2 - 6 (x - 1 x ) + C = 0 , where C is a real constant, has exactly two real roots, then the value of C is

Step-by-step solution

Let t = x - 1 x . Squaring both sides, we get t^2 = x^2 + 1 x^2 - 2 x^2 + 1 x^2 = t^2 + 2 . Substituting these into the given equation, we obtain: (t^2 + 2) - 6t + C = 0 t^2 - 6t + (C + 2) = 0 For any real value of t , the equation x - 1 x = t x^2 - tx - 1 = 0 has a discriminant D = t^2 + 4 . Since t^2 + 4 > 0 for all real t , every real root t yields exactly two distinct real roots for x . Therefore, for the original equation to have exactly two real roots, the quadratic equation in t must have exactly one real ro