It is a celebrated mathematical theorem that a formula exists which can solve general quartic equations. The formula consists of additions, subtractions, multiplications, divisions, and extraction of n th roots.

Consider the arbitrary quartic equation \ [ ax^4 + bx^3 + cx^2 + dx + e = 0 \] for real numbers $a$, $b$, $c$, $d$, $e$ with $a\neq0$. By the fundamental theorem of algebra this equation has four roots …

sgn 9bcd + 27ad2 + 27b2e 72ace)2 4(c2 3bd + 12ae)3 12a The quartic formula gives the solutions of ax4

The trick is that the $y_i$ were carefully chosen so that $y_1^2+y_2^2+y_3^2$, $y_1^2y_2^2 + y_1^2y_3^2 + y_2^2y_3^2$, and $y_1y_2y_3$ are symmetric polynomials in ...

The quartic formula gives the solutions of a x 4 + b x 3 + c x 2 + d x + e = 0 for real numbers a, b, c, d, e with a ≠ 0. Directions: Choose all possibilities for the three ± signs with the last two equivalent. Use …

4ac x = 2a The quadratic formula gives the solutions of ax2 + bx + c = 0 for real numbers a, b, c with a 6= 0.

p( 2b3 + 9abc 27a2d)2 4(b2 3ac)3 x = 6a The cubic formula gives the solutions of 3 ax + 2 bx + cx + d

The Quartic Formula The quartic formula gives the solutions of ax4 + bx3 + cx2 + dx + e = 0 for real numbers a, b, c, d, e with a ≠ 0. Directions: Choose all possibilities for the three ± signs with the last …

The quartic formula gives the solutions of $ax^4+bx^3+cx^2+dx+e=0$ for real numbers $a$, $b$, $c$, $d$, $e$ with $a\neq0$. Directions: Choose all possibilities for the three $\pm$ signs with the last …

\] From which the roots $y_1^3$ and $y_2^3$ can be found using the quadratic formula. Taking cube roots gives $y_1$ and $y_2$, although only 3 of the 9 possible choices for cube roots will work …