Solve ax^4+bx^3+cx^2+dx+e=0?
1 Answer
A quick sketch...
Explanation:
Given:
ax^4+bx^3+cx^2+dx+e = 0" " witha != 0
This gets messy quite quickly, so I will just give a sketch of one method...
Multiply by
t^4+pt^2+qt+r = 0
Note that since this has no term in
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)^3+2p(A^2)^2+(p^2-4d)(A^2)-q^2 = 0
This "cubic in
Given the value of
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.