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Germany | A nice Math Olympiad Algebra | Can you solve ?
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21 авг 2024

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Комментарии : 11

@MalikHamidov Месяц назад

Very very very goood and dificult question

@superacademy247 Месяц назад

Thanks and welcome. Keep watching!

@MalikHamidov Месяц назад

@@superacademy247 always 👍

@vacuumcarexpo Месяц назад

If you multiply by 9 both hands of the equation, you can get 3x(3x+4)(3x+8)(3x+12)=144. Considering the identity: x(x+a)(x+2a)(x+3a)+a^4=(x^2+3ax+a^2)^2, if you add 256 to both hands of this equation, you can get (9x^2+36x+16)^2=400 ⇔9x^2+36x-4=0 or 9x^2+36x+36=0 ⇔x=(-6±2√10)/3, -2

@SidneiMV Месяц назад

@@vacuumcarexpo awesome idea!

@SidneiMV Месяц назад

u = x + 2 => x = u - 2 (u - 2)(u + 2)(3u - 2)(3u + 2) = 16 (u² - 4)(9u² - 4) = 16 9u⁴ + 16 - 40u² = 16 9u⁴ - 40u² = 0 u²(9u² - 40) = 0 u = 0 => *x = -2* 9u² - 40 = 0 => u = ± (2√10)/3 => *x = -2 ± (2√10)/3*

@SidneiMV Месяц назад

another way x(3x + 4)(3x + 8)(x + 4) = 16 3x(3x + 4)(3x + 8)(3x + 12) = 144 3x + 6 = u => x = (u - 6)/3 3x(3x + 4)(3x + 8)(3x + 12) = 144 (u - 6)(u - 2)(u + 2)(u + 6) = 144 (u² - 36)(u² - 4) = 144 u⁴ - 40u² + 144 = 144 u²(u² - 40) = 0 u = 0 => x = -6/3 => *x = -2* u² = 40 => u = ± 2√10 x = (-6 ± 2√10)/3 *x = -2 ± (2√10)/3*

@pietergeerkens6324 Месяц назад

I did exactly the same, even to using "u" as the new variable. Very well done!

@walterwen2975 Месяц назад

A nice Math Olympiad Algebra: x(3x + 4)(3x + 8)(x + 4) = 16; x = ? x ϵR 9[x(3x + 4)(3x + 8)(x + 4)] = 9(16), (3x)(3x + 4)(3x + 8)(3x + 12) = 144 [(3x)(3x + 12)][(3x + 4)(3x + 8)] = (9x² + 36x)(9x² + 36x + 32) = 144 (9x² + 36x)² + 32(9x² + 36x) - 144 = 0, (9x² + 36x + 36)(9x² + 36x - 4) = 0 9x² + 36x + 36 = 0, x² + 4x + 4 = 0 or 9x² + 36x - 4 = 0, x² + 4x - 4/9 = 0 (x + 2)² = 0, x = - 2 or (x + 2)² = 40/9 = [(2√10)/3]², x = (- 6 ± 2√10)/3 Answer check: x(3x + 4)(3x + 8)(x + 4) = x(x + 4)(3x + 4)(3x + 8) = 16 x = - 2: (- 2)(- 6 + 4)(- 6 + 8)(- 2 + 4) = (- 2)(- 2)(2)(2) = 16; Confirmed x(x + 4) = [(- 6 ± 2√10)/3][(- 6 ± 2√10)/3 + 4] = [(- 6 ± 2√10)/3][(6 ± 2√10)/3 = (40 - 36)/9 = (4/9)(- 4 + 40) [(3x + 4)(3x + 8)] = (- 6 ± 2√10 + 4)(- 6 ± 2√10 + 8) = (- 2 ± 2√10)(2 ± 2√10) = 40 - 4 = 36, x(x + 4)(3x + 4)(3x + 8) = (4/9)(36) = (4)(4) = 16; Confirmed Final answer: x = - 2; Double root, x = (- 6 + 2√10)/3 or x = (- 6 - 2√10)/3