A Very Nice Math Olympiad Problem | Solve for x | Algebra | Exponential equation

[@renebrienne1862]

Simplify by X^2, then by x^2-1 : ok, but you have immediately précise, thé solution (s) MUST bé différent from , respectively : 0, +1, -1 (even if it IS quasi évident....)

[@Мефистофель-б6у]

Майнус и плайнус??

[@AliHassan-hb1bn]

Sub x with different variable, it is easier.

[@JamesKang95]

x^2=t t^4-t=9(t^2-t) t^4-9t^2+8t=0 t^3-9t+8=0 (t-1)(t^2+t-8)=0 t=1(no),t^2+t-8=0

[@oahuhawaii2141]

Close, but no cigar.

[@key_board_x]

(x⁸ - x²) / (x⁴ - x²) = 9 x².(x⁶ - 1) / [x².(x² - 1)] = 9 → where: x ≠ 0 and where: x ≠ ± 1 (x⁶ - 1) / (x² - 1) = 9 x⁶ - 1 = 9.(x² - 1) (x²)³ - 1³ = 9.(x² - 1) → recall: a³ - b³ = (a - b).(a² + ab + b²) (x² - 1).(x⁴ + x² + 1) = 9.(x² - 1) (x² - 1).(x⁴ + x² + 1) - 9.(x² - 1) = 0 (x² - 1).[(x⁴ + x² + 1) - 9] = 0 → recall: x ≠ ± 1 [(x⁴ + x² + 1) - 9] = 0 x⁴ + x² - 8 = 0 x⁴ + x² = 8 x⁴ + x² + (1/2)² = 8 + (1/2)² [x² + (1/2)]² = 33/4 x² + (1/2) = ± (√33)/2 x² = - (1/2) ± (√33)/2 x² = (- 1 ± √33)/2 First case: x² = (- 1 + √33)/2 x = ± √[(- 1 + √33)/2] Second case: x² = (- 1 - √33)/2 ← this is a negative value x² = - (1 + √33)/2 x² = i².(1 + √33)/2 x = ± i.√[(1 + √33)/2]

[@SpencersAcademy]

Excellent approach 👏

[@prasadrasikawidanagamachch3932]

X=[(+/-)√{-1+/-√33/2}]

[@oahuhawaii2141]

You're missing a pair of parentheses. Also, you're taking the square root of a negative number, and should convert it to standard form.