How do you find all the zeros of f(x) = 5x^4 - x^2 + 2f(x)=5x4−x2+2?
2 Answers
x = +-(sqrt((2sqrt(10)+1)/20)) +- (sqrt((2sqrt(10)-1)/20))ix=±⎛⎝√2√10+120⎞⎠±⎛⎝√2√10−120⎞⎠i
Explanation:
Since
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
Explanation:
I will use the result I derived for another question (see https://socratic.org/s/aw38evei), that the square roots of
+-(sqrt((sqrt(a^2+b^2)+a)/2)) +- (sqrt((sqrt(a^2+b^2)-a)/2))i
First treat
x^2=(1+-sqrt((-1)^2-4(5)(2)))/(2*5) = (1+-sqrt(-39))/10 = 1/10+-sqrt(39)/10i
Let
Then:
sqrt(a^2+b^2) = sqrt(1/100+39/100) = sqrt(40/100) = (2sqrt(10))/10
So the square roots of
+-(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.
