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[@key_board_x]

(1/x)² - (1/x)³ = 12 → where: x ≠ 0 (1/x)² - (1/x)³ = 12 → let: m = 1/x m² - m³ = 12 4 - (- 8) = 12 (- 2)² - (- 2)³ = 12 ← you can see that (m = - 2) is an obvious solution m² - m³ = 12 m³ - m² + 12 = 0 → as (m = - 2) is a solution, you can factorize: (m + 2) (m + 2).(m² + am + 6) = 0 → you expand m³ + am² + 6m + 2m² + 2am + 12 = 0 → you group m³ + (a + 2).m² + (6 + 2a).m + 12 = 0 → you compare with: m³ - m² + 12 = 0 For m² → (a + 2) = - 1 → a = - 3 For m → (6 + 2a) = 0 → 2a = - 6 → a = - 3 (m + 2).(m² + am + 6) = 0 → where: a = - 3 (m + 2).(m² - 3m + 6) = 0 First case: m = - 2 Recall: m = 1/x → x = - 1/2 Second case: m² - 3m + 6 = 0 Δ = (- 3)² - (4 * 6) = 9 - 24 = - 15 = 15i² m = (3 ± i√15)/2 → recall: m = 1/x x = 2/(3 ± i√15) First possibility: x = 2/(3 + i√15) x = 2/(3 + i√15) x = 2.(3 - i√15)/[(3 + i√15).(3 - i√15)] x = 2.(3 - i√15)/[(3)² - (i√15)²] x = 2.(3 - i√15)/[9 - 15i²] x = 2.(3 - i√15)/24 → x = (3 - i√15)/12 Second possibility: x = 2/(3 - i√15) x = 2/(3 - i√15) x = 2.(3 + i√15)/[(3 - i√15).(3 + i√15)] x = 2.(3 + i√15)/[(3)² - (i√15)²] x = 2.(3 + i√15)/[9 - 15i²] x = 2.(3 + i√15)/24 → x = (3 + i√15)/12 Summarize: x = - 1/2 x = (3 - i√15)/12 x = (3 + i√15)/12

[@johnbrennan3372]

(m+2)(m^2-3m+6) when you factorize m^3-m^2+12 knowing that m+2 is a factor(by inspection) where m=1\x. etc..

[@iqbalaulia6724]

if 1/x = a so a^2-a^3-12=0 a (a^2-a-12] =0 a=0 a 4 a =-3 x = 1/4 or 1-/3

[@superacademy247]

Thanks for sharing your insights. I'll have to try this out! 🥂🎁🎉

[@boguslawszostak1784]

a (a^2-a-12] =0 ????????????????????????????????????????

[@yorksters1]

Multiply both sides of equation by x^2. Write in standard form 12x^3-x+1 =0. Use rational root theorem to get list of possible rational roots: + or - (1,1/2,1/3,1/4,1/6,1/12). By synthetic division(use the short cut method) to recognize -1/2 is a root. Use the resulting quadratic root set = to 0 and solve for the remaining roots using the quadratic formula. Much quicker.

[@1234larry1]

Unless you’re in love with fractions why not just clear the fraction in the quadratic by multiplying through with x^2, which yields 6^x^2-3x+1=0?

[@1234larry1]

I'm not sure how you got a fraction inside the radical by I got (3+/-sqrt15i)/12

[@EC4U2C_Studioz]

Not familiar with the other quadratic formulas and would only use the most popular quadratic formula