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

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[@echandler]

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

[@walterwen2975]

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

[@RealQinnMalloryu4]

(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)