## Question

## Answers

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.