How do you solve x^4-sqrt(3)x^2+1= 0 ?
2 Answers
Explanation:
Note: all the three sets of answers are equivalent.
Making
So
Since
Now since
and
y=1/_color(white)()+-30^@
As
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:
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
Here are all of the
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) = 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