Question

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

Answer 1

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)`

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`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)`