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

2 Answers
Rui D.
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#

George C.
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 #12#th 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.

Related questions
Impact of this question
1181 views around the world
You can reuse this answer
Creative Commons License