#### AKG

Science Advisor

Homework Helper

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x

^{4}+ 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:

u

^{4}- 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

u

^{4}+ 3/4u² - 1/2u + 1/16 - 3u² + 3/2u - 1/4 + 3u² - 3u + 3/4 + 4u - 2 + 5 = 0

u

^{4}+ 3/4u² + 57/16 = 0

16u

^{4}+ 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:

[tex]x = \frac{1}{2} \pm \sqrt{\frac{-12 \pm \sqrt{-3504}}{32}}}[/tex]

So:

[tex]x = \frac{1 \pm \sqrt{\frac{-3 \pm \sqrt{-219}}{2}}}{2}[/tex]