A awesome mathematics problem | Olympiad Question | can you solve this problem | x=? 

after multiplying both sides by 16x^2, we have a^2 + b^2 = 4x^2 where a = x^2 - 20 - x b = x^2 - 20 + x now a^2 + b^2 = (a-b)^2 + 2ab and a-b = -2x, hence 4x^2 + 2ab = 4x^2 thus ab = 0 thus the roots are the roots of the individual quadratics a and b: 5, -4, -5, 4

Olympiad Question: [(x - 5)(x + 4)/4x]² + [(x + 5)(x - 4)/4x]² = 1/4; x = ? x ≠ 0; [(x - 5)(x + 4)]² + [(x + 5)(x - 4)]² = (1/4)[(4x)²] (x² - x - 20)² + (x² + x - 20)² = 4x², (x² - 20 - x)² + (x² - 20 + x)² = 4x² 2(x² - 20)² + 2x² = 4x², (x² - 20)² - x² = 0, (x² - x - 20)(x² + x - 20) = 0 (x - 5)(x + 4)(x - 4)(x + 5) = 0, x - 5 = 0; x + 4 = 0; x - 4 = 0 or x + 5 = 0 x = 5; x = - 4; x = 4 or x = - 5 Answer check: [(x - 5)(x + 4)/4x]² + [(x + 5)(x - 4)/4x] = 1/4 x = 5: 0 + [(5 + 5)(5 - 4)/20)]² = (1/2)² = 1/4; Confirmed x = 4: [(4 - 5)(4 + 4)/16]² + 0 = (- 1/2)² = 1/4; Confirmed x = - 4: 0 + [(- 4 + 5)(- 4 - 4)/(- 16)]² = (1/2)² = 1/4; Confirmed x = - 5: [(- 5 - 5)(- 5 + 4)/(- 20)]² + 0 = (- 1/2)² = 1/4; Confirmed Final answer: x = 5, x = 4, x = - 4 or x = - 5

(x ➖5x+1x)(x ➖2x+2)/2^2x)^2^2 ((x ➖ 1^1x +1^1)(x ➖ 1^1x+1^1)/1^1)1^2 (x ➖ 2x+1) (x ➖ 5x+1)(x ➖ 2x+2)/2^2)^2^2 (x ➖ 1^1x+1^1)(x ➖ 1^1x+1^1)/1^1)1^2 1^2 (x ➖ 2x+1)