The sum of all distinct real roots of the equation (4 4^ x^2 - 2x - 17 2^ x^2 - 2x + 4)(2 4^ x^2 - 2x - 17 2^ x^2 - 2x + 8) = 0 is

Step-by-step solution

Let y = 2^ x^2 - 2x . Then 4^ x^2 - 2x = y^2 . The equation becomes: (4y^2 - 17y + 4)(2y^2 - 17y + 8) = 0 Solving the first quadratic factor: 4y^2 - 17y + 4 = 0 4y^2 - 16y - y + 4 = 0 4y(y - 4) - 1(y - 4) = 0 (4y - 1)(y - 4) = 0 y = 1 4 or y = 4 Solving the second quadratic factor: 2y^2 - 17y + 8 = 0 2y^2 - 16y - y + 8 = 0 2y(y - 8) - 1(y - 8) = 0 (2y - 1)(y - 8) = 0 y = 1 2 or y = 8 Now, we find the real roots for x from 2^ x^2 - 2x = y , which means x^2 - 2x = ₂ y . Case 1: y = 4 x^2 - 2x = 2 x^2 - 2x - 2 = 0 Dis