How can this polynomial be solved?

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x^4 - 4x^2 - 1
= x^2(x^2 - 4) - 1
This has 2 real and 2 complex roots....Show more

It is a "biquadratic".

If you replace x^2 by another variable (let's use "u"), it then becomes a quadratic

u = x^2

u^2 - 4u - 1
is the same as x^4 - 4x^2 - 1

Using the quadratic formula:

u = [ +4 +/- SQRT( 16 + 4 ) ] / 2
u = [ 4 +/- √(20) ] /2
u = (4 +/- 4√5)/2
u = 2 +/- √5

One possible solution is
u = 2 + √5
the other possible solution is
u = 2 - √5

I say "possible" because u is not what we are looking for; we want x

u = x^2
therefore
x = +/- √u

The first possible pair of solutions is

x = +/-√(2 + √5)
Yes, there is a root inside a root. This looks like a weird beast, but it is "simply" a number.
The approximate value is 2.058171...
x can be the positive version (+2.058171...) and it can also be the negative version (-2.058171...)

The second possible pair of solutions would be:

x = +/-√(2 - √5)
However, √5 is bigger in value than 2, therefore we would be looking for the square root of a negative number. If you are working in real numbers, then this pair of solutions will NOT exist.

If you are allowed to use "imaginary numbers", the pair of solutions would be
x = +/- i√(√5 - 2) = approx. +/- i(0.48586827...)
where "i" is the imaginary number, such that i^2 = -1
(it is a number that was "imagined" as a square root of -1)...Show more

I am very bad at math....Show more

(x^2 -2-\/''''5'''')(x^2 -2 +\/'''''5''')
x^2=+\/''''5''' +2
&
x^2=\/'''''5'''' -2
God bless you....Show more