An Amazing Radical Challenge | Math Olympiad Prep

[@woobjun2582]

Rearranging the given, (x +12/x +1)/x = √(1/x +12/x² +3); 1/x +12/x² +1 = √(1/x +12/x² +1 +2). where x ≠0. Letting t = 1/x +12/x² +1, t = √(t +2) in which t > 0. And squaring, t²= t +2; t² -t -2 =0; t² -2t +t -2 =0; t(t -2) +(t -2) =0; (t -2)(t +1) =0, that is t = 2 t = -1 (rejected) Thus, t = 2 = 1/x +12/x² +1; 1/x +12/x² -1 =0, and then multiplying -x², x² -x -12 =0; x² -4x +3x -12 =0; x(x -4) +3(x -4) =0; (x -4)(x +3) =0, that is, x = 4, -3

[@paulortega5317]

Let u = 1/x + 1/x^2 +3 (u - 2) = sqrt(u) u = 1 or 4 for u = 1 and v = 1/x, 12*v^2 + v + 2 = 0, complex solutions only for u = 4 and v = 1/x, 12*v^2 + v - 1 = 0, v = -1/3 or 1/4, x = -3 or 4

[@mohammedsaysrashid3587]

It was a wonderful explanation ,thank you Sir for sharing x=-3,4

[@infyGyan]

Glad you liked it Thanks

[@Quest3669]

4; -3

[@gregevgeni1864]

(x+3)(x-4)(2x^2 +x+12)=0 etc ..