A Very Nice Math Olympiad Problem | Solve for x? | Algebra Equation

[@dudasslobodan8901]

X=+-√11i
X=+-✓11√(-1)
,√(-1)=i
X=+-√11i NO. +-√-11i

[@bandarusatyanandachary1181]

The equation satisfy at x = 0

[@walterwen2975]

A Very Nice Math Olympiad Problem:
[(x + 2)(x + 3)(x + 4)(x + 5)]/[(x - 2)(x - 3)(x - 4)(x - 5)] = 1; x =?
[(x + 2)(x + 5)][(x + 3)(x + 4)] = (x² + 7x + 10)(x² + 7x + 12)
= (x² + 10 + 7x)(x² + 12 + 7x) = (x² + 10)(x² + 12) + 14x(x² + 11) + (7x)²
[(x - 2)(x - 5)][(x - 3)(x - 4)] = (x² - 7x + 10)(x² - 7x + 12)
= (x² + 10 - 7x)(x² + 12 - 7x) = (x² + 10)(x² + 12) - 14x(x² + 11) + (7x)²
(x + 2)(x + 3)(x + 4)(x + 5) = (x - 2)(x - 3)(x - 4)(x - 5)
(x² + 10)(x² + 12) + 14x(x² + 11) + (7x)² = (x² + 10)(x² + 12) - 14x(x² + 11) + (7x)²
28x(x² + 11) = 0, x(x² + 11) = 0; x = 0 or x² + 11 = 0, x² = - 11; x = ± i√11
Answer check:
[(x + 2)(x + 3)(x + 4)(x + 5)]/[(x - 2)(x - 3)(x - 4)(x - 5)]
= [(x² + 7x + 10)(x² + 7x + 12)]/[(x² + 7x + 10)(x² + 7x + 12)] = 1
x = 0: [(2)(3)(4)(5)]/[(- 2)(- 3)(- 4)(- 5)] = 1; Confirmed
x = ± i√11: x² = - 11
x² + 7x + 10 = - 11 ± 7i√11 + 10 = - 1 ± 7i√11, x² + 7x + 12 = 1 ± 7i√11
(x² + 7x + 10)(x² + 7x + 12) = (- 1 ± 7i√11)(1 ± 7i√11) = - 1 - 539 = - 540
x² - 7x + 10 = - 1 -/+ 7i√11, x² - 7x + 12 = 1 -/+ 7i√11
(x² - 7x + 10)(x² - 7x + 12) = (- 1 -/+ 7i√11)(1 -/+ 7i√11) = - 1 - 539 = - 540
[(x² + 7x + 10)(x² + 7x + 12)]/[(x² + 7x + 10)(x² + 7x + 12)] = 1; Confirmed
Final answer:
x = 0; x = i√11 or x = - i√11