This is one possible way to solve this problem. For my purposes, I will use the symbols #a#, #b#, #c#, and #d# to represent the four roots.
From the original statement, we know that #a + b = 0#. This means that #x = a# and #x = -a# are roots of the equation. Furthermore, that means that #f(a) = 0# and #f(-a) = 0#, since roots of equations are also places where the equation evaluates to a value of 0. If we evaluate #a# and #-a# into the original equation, we can derive a set of two equations which we can solve simultaneously:
#f(a) = a^4 - 2a^3 + 4a^2 + 6a - 21 = 0#
#f(-a) = a^4 + 2a^3 + 4a^2 - 6a - 21 = 0#
Since both are equal to 0, we can add each together to get a new equation which also equals 0:
#f(a) + f(-a) = 0 =>#
#2a^4 + 8a^2 - 42 = 0#
#a^4 + 4a^2 - 21 = 0#
This last equation we can further breakdown by considering the substitution #w = a^2# and rewriting:
#w^2 + 4w - 21 = 0#
Solving this with the Quadratic Formula yields possible values for #w#:
#w = (-4 +-sqrt(4^2-4*1*(-21)))/2 = (-4 +- sqrt(100))/2 #
#w = (-4 +- 10)/2 = 3 or -7#
Recalling that we had substituted #w = a^2#, we can finally work out the value(s) of #a#:
#w = a^2 #
#{(a^2 = 3, a = +-sqrt(3)), (a^2 = -7, "No solutions"):}#
We have two of the roots found: #x = sqrt(3), x = -sqrt(3)#. Noting that these represent the factors #(x-sqrt(3))(x + sqrt(3))#, we can now divide the original equation by these two factors (the work of which I will leave to the reader), and we reduce the equation down to the following:
#x^2 - 2x + 7#
This can also be solved using the Quadratic Formula:
#x = (2 +- sqrt((-2)^2 - 4*1*7))/2 = (2 +- sqrt(-24))/2 #
# = (2 +- 2isqrt(6))/2 = 1 +- isqrt(6)#
Our final two roots are #x = 1 + isqrt(6), x = 1 - isqrt(6)#.
