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

1 Answer
Mar 18, 2018

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.