Question

Solve `ax^4+bx^3+cx^2+dx+e=0`?

Answer 1

A quick sketch...

Explanation:

Given:

`ax^4+bx^3+cx^2+dx+e = 0" "` with `a != 0`

This gets messy quite quickly, so I will just give a sketch of one method...

Multiply by `256a^3` and substitute `t = (4ax+b)` to get a depressed monic quartic of the form:

`t^4+pt^2+qt+r = 0`

Note that since this has no term in `t^3`, it must factor in the form:

`t^4+pt^2+qt+r = (t^2-At+B)(t^2+At+C)`

`color(white)(t^4+pt^2+qt+r) = t^4+(B+C-A^2)t^2+A(B-C)t+BC`

Equating coefficients and rearranging a little, we have:

`{ (B+C = A^2+p), (B-C = q/A), (BC = d) :}`

So we find:

`(A^2+p)^2 = (B+C)^2`

`color(white)((A^2+p)^2) = (B-C)^2 + 4BC`

`color(white)((A^2+p)^2) = q^2/A^2 + 4d`

Multiplying out, multiplying by `A^2` and rearranging slightly, this becomes:

`(A^2)^3+2p(A^2)^2+(p^2-4d)(A^2)-q^2 = 0`

This "cubic in `A^2`" has at least one real root. Ideally it has a positive real root that yields two possible real values for `A`. Regardless, any root of the cubic will do.

Given the value of `A`, we have:

`B = 1/2((B+C)+(B-C)) = 1/2(A^2+p+q/A)`

`C = 1/2((B+C)-(B-C)) = 1/2(A^2+p-q/A)`

Hence we get two quadratics to solve.