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
@SidneiMV3 ай бұрын
@@vacuumcarexpo awesome idea!
@MalikHamidov3 ай бұрын
Very very very goood and dificult question
@superacademy2473 ай бұрын
Thanks and welcome. Keep watching!
@MalikHamidov3 ай бұрын
@@superacademy247 always 👍
@SidneiMV3 ай бұрын
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*
@SidneiMV3 ай бұрын
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*
@pietergeerkens63243 ай бұрын
I did exactly the same, even to using "u" as the new variable. Very well done!