by AKG
 Sci Advisor HW Helper P: 2,586 I missed the lectures for this topic, so I don't have the notes, so I was wondering if anyone could give me the idea behind how to solve quartics in radicals. I know its long and messy, so just the basic idea would do. For example: x4 + 2x³ + 3x² + 4x + 5 = 0 I recall something about getting rid of the cubic term, so maybe I should substitute x = (u - 1/2), giving: u4 - 2u³ + 3/4u² - 1/2u + 1/16 + 2(u³ - 3/2u² + 3/4u - 1/8) + 3(u² - u + 1/4) + 4(u - 1/2) + 5 = 0 u4 + 3/4u² - 1/2u + 1/16 - 3u² + 3/2u - 1/4 + 3u² - 3u + 3/4 + 4u - 2 + 5 = 0 u4 + 3/4u² + 57/16 = 0 16u4 + 12u² + 57 = 0 Okay, this one happened to work out nicely, with the degree-1 term going away as well. Now I just have a quadratic in u². So perhaps a different example would be more enlightening. Also, when asked to "solve in radicals" does that mean that the correct answer to the above problem should be given as: $$x = \frac{1}{2} \pm \sqrt{\frac{-12 \pm \sqrt{-3504}}{32}}}$$ So: $$x = \frac{1 \pm \sqrt{\frac{-3 \pm \sqrt{-219}}{2}}}{2}$$