How do you solve x^4-sqrt(3)x^2+1= 0 ?

2 Answers
Nov 2, 2017

(+-sqrt((sqrt3+i)/2),+-sqrt((sqrt3-i)/2)) or (cos 15^@+i*sin 15^@, cos 15^@-i*sin 15^@,-cos 15^@+i*sin 15^@,-cos 15^@-i*sin 15^@) or (1/_15^@,1/_165^@,1/_color(white)()-15^@,1/_color(white)()-165^@)

Explanation:

Note: all the three sets of answers are equivalent.

Making y=x^2 we get:

y^2-sqrt3*y+1=0
Delta=b^2-4ac=(-sqrt3)^2-4*1*1=3-4=-1
So
y=(-b+-sqrt(Delta))/(2a)=(sqrt3+-i)/2
Since x=sqrt(y)
x=+-sqrt((sqrt(3)+-i)/2) -> it means 4 roots

Now since |y|=sqrt((sqrt3/2)^2+(1/2)^2)=1
and arctan((1/cancel(2))/(sqrt3/cancel(2)))=arctan(1/sqrt3)=30^@
y can also be written in this way (polar form):

y=1/_color(white)()+-30^@

As x=sqrt(y), we have 4 roots:

x'=sqrt1/_+-30^@/2=1/_color(white)()+-15^@
x''=sqrt1/_color(white)()+-(30^@+360^@)/2=1/_color(white)()+-195^@=1/_color(white)()+-165^@

Finally these 4 mentioned roots can be written in this way:

x_1=cos 15^@+i*sin 15^@
x_2=cos (-15^@)+i*sin (-15^@)=cos 15^@-i*sin 15^@
x_3=cos 165^@+i*sin 165^@=-cos 15^@ +i*sin 15^@
x_4=cos (-165^@)+i*sin (-165^@)=-cos 15 -i*sin 15

Nov 2, 2017

x = +-1/4(sqrt(6)+sqrt(2))+-1/4(sqrt(6)-sqrt(2))i

Explanation:

Given:

x^4-sqrt(3)x^2+1= 0

I really like this question, for reasons which may become apparent.

First look at this:

(x^4-sqrt(3)x^2+1)(x^4+sqrt(3)x^2+1) = x^8-x^4+1

Then:

(x^4+1)(x^8-x^4+1) = x^12+1

So the roots of x^4-sqrt(3)x^2+1 are four of the twelve twelfth roots of -1.

Here are all of the 12th roots of -1 in the complex plane:

graph{((x-cos(pi/12))^2+(y-sin(pi/12))^2-0.002)((x-cos(3pi/12))^2+(y-sin(3pi/12))^2-0.002)((x-cos(5pi/12))^2+(y-sin(5pi/12))^2-0.002)((x-cos(7pi/12))^2+(y-sin(7pi/12))^2-0.002)((x-cos(9pi/12))^2+(y-sin(9pi/12))^2-0.002)((x-cos(11pi/12))^2+(y-sin(11pi/12))^2-0.002)((x-cos(13pi/12))^2+(y-sin(13pi/12))^2-0.002)((x-cos(15pi/12))^2+(y-sin(15pi/12))^2-0.002)((x-cos(17pi/12))^2+(y-sin(17pi/12))^2-0.002)((x-cos(19pi/12))^2+(y-sin(19pi/12))^2-0.002)((x-cos(21pi/12))^2+(y-sin(21pi/12))^2-0.002)((x-cos(23pi/12))^2+(y-sin(23pi/12))^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}

In fact note that:

x^4-sqrt(3)x^2+1 = (x^2)^2-2cos(pi/6)(x^2)+1

color(white)(x^4-sqrt(3)x^2+1) = (x^2-cos(pi/6)-isin(pi/6)))(x^2-cos(-pi/6)-isin(-pi/6))

So we find:

x^2 = cos(pi/6)+isin(pi/6)" " or " "x^2 = cos(-pi/6)+isin(-pi/6)

Hence by de Moivre's formula:

x = +-(cos(pi/12)+isin(pi/12))

or:

x = +-(cos(-pi/12)+isin(-pi/12))

So the four roots of our quartic are these ones:

graph{((x-cos(pi/12))^2+(y-sin(pi/12))^2-0.002)((x-cos(11pi/12))^2+(y-sin(11pi/12))^2-0.002)((x-cos(13pi/12))^2+(y-sin(13pi/12))^2-0.002)((x-cos(23pi/12))^2+(y-sin(23pi/12))^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}

Note also that we can get exact algebraic formulae for cos(pi/12), sin(pi/12), etc. as follows:

cos(pi/12) = cos(pi/3 - pi/4)

color(white)(cos(pi/12)) = cos(pi/3) cos(pi/4) + sin(pi/3) sin(pi/4)

color(white)(cos(pi/12)) = 1/2 sqrt(2)/2 + sqrt(3)/2 sqrt(2)/2

color(white)(cos(pi/12)) = 1/4 (sqrt(6)+sqrt(2))

sin(pi/12) = sin(pi/3-pi/4)

color(white)(sin(pi/12)) = sin(pi/3)cos(pi/4)-sin(pi/4)cos(pi/3)

color(white)(sin(pi/12)) = sqrt(3)/2 sqrt(2)/2-sqrt(2)/2 1/2

color(white)(sin(pi/12)) = 1/4 (sqrt(6)-sqrt(2))

So the roots of our quartic can be written:

x = +-1/4(sqrt(6)+sqrt(2))+-1/4(sqrt(6)-sqrt(2))i

where any combination of +- signs is included.