How do you solve #x^4 + 2x + 2 = 0#?

1 Answer
Aug 11, 2016

Answer:

See explanation for an exploration of solving this quartic algebraically...

Explanation:

#f(x) = x^4+2x+2#

Looks quite simple, does it not?

Solving quartics algebraically can get rather messy, even for an example as simple as this, but let's give it a try...

Since #f(x)# is monic and has no cubic term, it can be factored as a product of monic quadratics with opposite middle coefficients:

#x^4+2x+2#

#=(x^2-ax+b)(x^2+ax+c)#

#=x^4+(b+c-a^2)x^2+(b-c)ax+bc#

Equating coefficients and rearranging a little we have:

#{ (b+c = a^2), (b-c=2/a), (bc=2) :}#

Hence we find:

#(a^2)^2 = (b+c)^2 = (b-c)^2+4bc = 4/((a^2))+8#

Multiplying both ends by #(a^2)# and rearranging slightly, we get a cubic in #(a^2)#:

#(a^2)^3-8(a^2)-4 = 0#

This cubic has #3# Real zeros, so if you try to solve it using Cardano's method you get solutions involving irreducible cube roots of Complex numbers (casus irreducibilis).

I will address this cubic in a separate question (see https://socratic.org/s/awX2qsH7), but suffice it to say that we are interested in the positive root (which I will call #r_3# here):

#r_3 = 4/3 sqrt(6) cos(1/3 cos^(-1)(3/16 sqrt(6))) ~~ 3.05#

Let #a=sqrt(r_3)#

Then:

#b=1/2(a^2+2/a) = r_3/2+1/sqrt(r_3)#

#c=1/2(a^2-2/a) = r_3/2-1/sqrt(r_3)#

That leaves us with two quadratics to solve:

#x^2-sqrt(r_3)x+(r_3/2+1/sqrt(r_3)) = 0#

#x^2+sqrt(r_3)x+(r_3/2-1/sqrt(r_3)) = 0#

Using the quadratic formula, we find roots:

#x_(1,2) = (sqrt(r_3)+-sqrt(r_3-4(r_3/2+1/sqrt(r_3))))/2#

#= (sqrt(r_3)+-sqrt(r_3-2r_3-4/sqrt(r_3)))/2#

#=1/2(sqrt(r_3)+-(sqrt(r_3+4/sqrt(r_3)))i)#

#color(white)()#

#x_(3,4) = (-sqrt(r_3)+-sqrt(r_3-4(r_3/2-1/sqrt(r_3))))/2#

#= (-sqrt(r_3)+-sqrt(r_3-2r_3+4/sqrt(r_3)))/2#

#=1/2(-sqrt(r_3)+-(sqrt(r_3-4/sqrt(r_3)))i)#