text: We have:\n(2+√3)^(x^2 - 2x + 1) + (2-√3)^(x^2 - 2x - 1) = 4/(2-√3)\n\n(2-√3) [(2+√3)^(x^2 - 2x + 1) + (2-√3)^(x^2 - 2x - 1)] = 4\n(2-√3)(2+√3)^(x^2 - 2x + 1) + (2-√3)(2-√3)^(x^2 - 2x - 1) = 4\n(2-√3)(2+√3)(2+√3)^(x^2 - 2x) + (2-√3)^(x^2 - 2x) = 4\nand (2-√3)(2+√3) = 1, so we have:\n\n(2+√3)^(x^2 - 2x) + (2-√3)^(x^2 - 2x) = 4\nagain, (2-√3)(2+√3) = 1 so (2-√3) = 1 /(2+√3) \nsubstituting,\n\n(2+√3)^(x^2 - 2x) + 1/(2+√3)^(x^2 - 2x) = 4\nLet y = (2+√3)^(x^2 - 2x)\nso y + 1/y = 4\nor y^2 - 4y + 1 = 0\ny can then be found to be 2 ± √3\n\nso (2+√3)^(x^2 - 2x) = 2 ± √3\nif (2+√3)^(x^2 - 2x) = 2 + √3, then x^2 - 2x = 1\nif (2+√3)^(x^2 - 2x) = 2 - √3 = (2 + √3)^(-1), then x^2 - 2x = -1\n\nsolving these for x, we find x = 1 ± √2 or the double root x = 1