A Nice Algebra Problem | Math Olympiad | A Nice Equation

[@فیروزاهنگری]

very nice❤

[@brianwade4179]

At 2:33 use the Rational Root Theorem to discover x=-1 as a root. Then use polynomial division to find x^2-x-3 as the other factor. Then complete the square to find the roots of the quadratic.

[@SALogics]

Very nice! ❤

[@stpat7614]

Slightly different method:
3^(x^3) / 81^x = 27
3^(x^3) / (3^4)^x = 3^3
3^(x^3) / 3^(4*x) = 3^3
3^(x^3 - 4*x) = 3^3
x^3 - 4x = 3
x^3 - 4x - 3 = 3 - 3
x^3 - 4x - 3 = 0
x^3 - (1x + 3x) - 1*3 = 0
(x^3 - 1*x) + (-3*x - 1*3) = 0
x(x^2 - 1) - 3(x + 1) = 0
x(x^2 - 1^2) - 3(x + 1) = 0
x(x - 1)(x + 1) - 3(x + 1) = 0
(x + 1)(x[x - 1] - 3) = 0
(x + 1)(x^2 - x - 3) = 0
x + 1 = 0, or 1*x^2 - 1*x - 3 = 0
Let a = 1, b = -1, c = -3
x + 1 = 0, or x = (-b +/- sqrt[b^2 - 4*a*c]) / (2*a)
x + 1 - 1 = 0 - 1, or x = (-[-1] +/- sqrt[(-1)^2 - 4*1*(-3)]) / (2*1)
x = - 1, or x = (1 +/- sqrt[1 + 12]) / (2)
x = - 1, or x = (1 +/- sqrt[13]) / 2
x = - 1, or x = (1 + sqrt[13]) / 2, or x = (1 - sqrt[13]) / 2
x1 = -1
x2 = (1 + sqrt[13]) / 2
x3 = (1 - sqrt[13]) / 2

[@SALogics]

Very nice! ❤

[@walterwen2975]

Math Olympiad: (3^x³)/(81ˣ) = 27; x =?
(3^x³)/(81ˣ) = (3^x³)/(3⁴ˣ) = 3^(x³ - 4x) = 27 = 3^3, x³ - 4x = 3, x³ - 4x - 3 = 0
(x³ + 1) - (4x + 4) = 0, (x + 1)(x² - x + 1) - 4(x + 1) = (x + 1)(x² - x - 3) = 0
x + 1 = 0, x = - 1 or x² - x - 3 = 0, x = (1 ± √13)/2
Answer check:
x = - 1: (3^x³)/(81ˣ) = (3⁻¹)/(81⁻¹) = 81/3 = 27; Confirmed
x = (1 ± √13)/2, x² - x - 3 = 0, x² = x + 3: x³ = x² + 3x = (x + 3) + 3x = 4x + 3
(3^x³)/(81ˣ) = 3^(x³ - 4x) = 3^(4x + 3 - 4x) = 3³ = 27; Confirmed
Final answer:
x = - 1; x = (1 + √13)/2 or x = (1 - √13)/2

[@SALogics]

Very nice! ❤