What are the possible solutions of #ax^4+bx^3+cx^2+dx+e=0# where #a!=0#, #a+c+e=0# and #b+d=0#?

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Answer 1 · 2018-03-22T07:09:03

This quartic has roots #x=+-1# and:

#x = (-b+-sqrt(b^2-4a(a+c)))/(2a)#

Let:

#f(x) = ax^4+bx^3+cx^2+dx+e#

with #a != 0#, #a+c+e = 0# and #b + d = 0#

Then:

#f(1) = a+b+c+d+e = (a+c+e)+(b+d) = 0#

#f(-1) = a-b+c-d+e = (a+c+e)-(b+d) = 0#

So #x=1# and #x=-1# are zeros with corresponding factors #(x-1)# and #(x+1)#.

Hence #f(x)# is divisible by:

#(x-1)(x+1) = x^2-1#

We have:

#ax^4+bx^3+cx^2+dx+e = (x^2-1)(ax^2+bx+(a+c))#

Note that from the given conditions on #a, b, c, d, e# there are no restrictions on #a, b, c# apart from #a != 0#.

So the remaining quadratic can have any coefficients.

We can express the solutions using the quadratic formula, substituting #(a+c)# for "c" to find zeros:

#x = (-b+-sqrt(b^2-4a(a+c)))/(2a)#