A awesome mathematics exponential problem | Olympiad Question | x=?

[@MuhammadJaveed-m9s]

Very nice

[@mircoceccarelli6689]

👍

[@NaeemKhan-q6w1v]

❤😂🎉😢😢😮😅😊

[@RealQinnMalloryu4]

(x ➖ 3x+2).

[@Mauuro-h4d]

x+(12/x)+1=x•sqrt[(1/x)+(12/x²)+3]
○Dividing by x both sides:
(1/x)+(12/x²)+1=sqrt[(1/x)+(12/x²)+3]
○Let a= (1/x)+(12/x²)+1:
a=sqrt(a+2) with a≥0
a²=a+2→a²-a-2=0
→a=2 or a=-1 (reject a=-1

[@walterwen2975]

Olympiad Question: (x + 12/x + 1)/√(1/x + 12/x² + 3) = x, x ≠ 0, x ϵ R; x =?
x(x + 12/x + 1) = (x²)√(1/x + 12/x² + 3) = √[(x⁴)(1/x + 12/x² + 3)] = √(3x⁴ + x³ + 12x²)
(x² + x + 12)² = (x² + x)² + 24(x² + x) + 144 = 3x⁴ + x³ + 12x²
x⁴ + 2x³ + 25x² + 24x + 144 = 3x⁴ + x³ + 12x², 2x⁴ - x³ - 13x² - 24x - 144 = 0
2x⁴ - x²(x + 12) - (x² + 24x + 144) = 2x⁴ - x²(x + 12) - (x + 12)² = 0
[x² - (x + 12)][2x² + (x + 12)] = 0, (x² - x - 12)(2x² + x + 12) = 0
2x² + x + 12 = 0, x = (- 1 ± i√95)/2, Rejected; x ϵ R
x² - x - 12 = 0, (x - 4)(x + 3) = 0; x - 4 = 0, x = 4 or x + 3, x = - 3
Answer check:
(x + 12/x + 1)/√(1/x + 12/x² + 3) = x
x = 4: (4 + 12/4 + 1)/√(1/4 + 12/16 + 3) = 8/√4 = 4; Confirmed
x = - 3: (- 3 - 12/3 + 1)/√(- 1/3 + 12/9 + 3) = - 6/√4 = - 3; Confirmed
Final answer:
x = 4 or x = - 3