A nice Algebra Problem | Math Olympiad | Quadratic Factorial Question | Can you solve this ?  

A nice but simple problem. Get us something really challenging

A nice Algebra Problem: x² - 9x = (√25)!/√(6!); x = ? x² - 9x = (√25)!/√(6!) = 5!/√720 = [5(4)(3)(2)]/[(12)(√5)] = [(2)(5)]/√5 = 2√5 x² - 9x - 2√5 = 0, ∆ = 81 - 4(- 2√5) = 81 + 8√5; x = [9 ± √(81 + 8√5)]/2 81 + 8√5 = 81 + 2√[(16)(5)] = 80 + 2√80 + 1 = (√80 + 1)² = (4√5 + 1)² √(81 + 8√5) = √[(4√5 + 1)²] = 4√5 + 1; x = [9 ± (4√5 + 1)]/2 x = (10 + 4√5)/2 or x = (8 - 4√5)/2 Answer check: x² - 9x = x(x - 9) = 2√5 x = (10 + 4√5)/2: x - 9 = 8(10 + 4√5)/2 - 9 = (- 8 + 4√5)/2 x² - 9x = [(10 + 4√5)/2][(- 8 + 4√5)/2] = (8√5)/4 = 2√5; Confirmed x = (8 - 4√5)/2: x - 9 = (8 - 4√5)/2 - 9 = (- 10 - 4√5)/2 x² - 9x = [(8 - 4√5)/2][(- 10 - 4√5)/2] = 8√5)/4 = 2; Confirmed Final answer: x = (10 + 4√5)/2 or x = (8 - 4√5)/2

What does "solutiun" mean?

Great algebraic exercise.