Question

How do you find all the zeros of `f(x) = 5x^4 - x^2 + 2`?

Answer 1

`x = +-(sqrt((2sqrt(10)+1)/20)) +- (sqrt((2sqrt(10)-1)/20))i`

Explanation:

Since `f(x)` has no odd terms, we could treat it as a quadratic in `x^2`, but the resulting quadratic has negative discriminant:

`Delta = (-1)^2-4(5)(2) = 1-40 = -39`

As a result it has only Complex zeros, of which we would then want to find square roots. This is a possible approach, but there is an alternative one...

Note that:

`(a^2-kab+b^2)(a^2+kab+b^2) = a^4 + (2-k^2)a^2b^2 + b^4`

Let:

`a = root(4)(5)x`

`b = root(4)(2)`

Then:

`a^4 + (2-k^2)a^2b^2 + b^4=5x^4+(2-k^2)sqrt(10)x^2+2`

If we let:

`k = sqrt((20+sqrt(10))/10)`

Then:

`(2-k^2)sqrt(10) = (2-(20+sqrt(10))/10)sqrt(10) = -1`

And:

`kab = sqrt((20+sqrt(10))/10)root(4)(10)x = sqrt((20+sqrt(10))/10*sqrt(10)) = sqrt(2sqrt(10)+1)`

So:

`f(x) = (a^2-kab+b^2)(a^2+kab+b^2)`

`=(sqrt(5)x^2-(sqrt(2sqrt(10)+1))x+sqrt(2))(sqrt(5)x^2+(sqrt(2sqrt(10)+1))x+sqrt(2))`

Then the zeros of these quadratic factors are given by the quadratic formula as:

`x = (+-sqrt(2sqrt(10)+1)+-sqrt((2sqrt(10)+1)-4sqrt(10)))/(2sqrt(5))`

`=+-(sqrt((2sqrt(10)+1)/20)) +- (sqrt((2sqrt(10)-1)/20))i`

Answer 2

`x=+-(sqrt((2sqrt(10)+1)/20)) +- (sqrt((2sqrt(10)-1)/20))i`

Explanation:

`f(x) = 5x^4-x^2+2`

I will use the result I derived for another question (see https://socratic.org/s/aw38evei), that the square roots of `a+-bi` are:

`+-(sqrt((sqrt(a^2+b^2)+a)/2)) +- (sqrt((sqrt(a^2+b^2)-a)/2))i`

First treat `f(x)` as a quadratic in `x^2` and use the quadratic formula to find:

`x^2=(1+-sqrt((-1)^2-4(5)(2)))/(2*5) = (1+-sqrt(-39))/10 = 1/10+-sqrt(39)/10i`

Let `a=1/10` and `b=sqrt(39)/10`

Then:

`sqrt(a^2+b^2) = sqrt(1/100+39/100) = sqrt(40/100) = (2sqrt(10))/10`

So the square roots of `a+bi` are:

`+-(sqrt((sqrt(a^2+b^2)+a)/2)) +- (sqrt((sqrt(a^2+b^2)-a)/2))i`

`=+-(sqrt(((2sqrt(10))/10+1/10)/2)) +- (sqrt(((2sqrt(10))/10-1/10)/2))i`

`=+-(sqrt((2sqrt(10)+1)/20)) +- (sqrt((2sqrt(10)-1)/20))i`

These are the zeros of our quartic.