Find the zeros of F(x)=x^4+x^2+1.?

Question

Find all the zeros, complex or real. Explain your answer please.

Answer 1

let y=x^2, then your equation reduces to:

y^2 + y +1 = 0

y = { -1 + sqrt(-3) } / 2

y = -1/2 + i*(√3)/2
y = -1/2 - i*(√3)/2

y = cos(120) + i*sin(120)
y = cos(120) - i*sin(120) y = cos(240) + i*sin(240)

We have got two solutions for 'y', now substitute 'x':

x^2 = cos(120) + i*sin(120)
x^2 = cos(240) + i*sin(240)

x = (cos(120) + i*sin(120)) ^ (1/2)
x = (cos(240) + i*sin(240)) ^ (1/2)

Using Euler's identity, we can say:

x = cos(120/2) + i*sin(120/2)
x = cos(240/2) + i*sin(240/2)

x = cos(60) + i*sin(60)
x = cos(120) + i*sin(120)

The above are the first set of roots for the equation. To get the other roots, keep adding 360 degrees to the angles until you cycle back to the first set of roots (as the concept goes, adding 360 degrees to an angle will not alter it's properties in any way):

x = (cos(120+360) + i*sin(120+360)) ^ (1/2)
x = (cos(240+360) + i*sin(240+360)) ^ (1/2)

x = cos(480/2) + i*sin(480/2)
x = cos(600/2) + i*sin(600/2)

x = cos(240) + i*sin(240)
x = cos(300) + i*sin(300)

These are the other set of roots, now if you add 360 degrees again, you will see that it cycles back to the first set of roots.


Hence, the roots or zeros of the equation are:

x = cos(60) + i*sin(60)
x = cos(120) + i*sin(120)
x = cos(240) + i*sin(240)
x = cos(300) + i*sin(300)

or, in terms of real numbers:

x = 1/2 + i*(√3)/2
x = -1/2 + i*(√3)/2
x = -1/2 - i*(√3)/2
x = 1/2 - i*(√3)/2

are the four zeros of the equation.

Answer 2

OK, notice that this equation is really a quadratic equation in disguise. That is, F(x) = (x^2)^2 + (x^2) + 1
So, apply the quadratic formula:
x^2 = (-1 (+-) sqrt(1-4))/2 = -1/2 + i*sqrt(3)/2 or -1/2 - i*sqrt(3)/2
therefore,
x = (+/-) sqrt(-1/2 + i*sqrt(3)/2) or
x = (+/-) sqrt(-1/2 - i*sqrt(3)/2).

You can confirm them by multiplying them out like this: (x-a)(x-b)(x-c)(x-d), where a, b, c, and d are the zeros above. Good luck!

Mathgirl: You're right about the quickmath website. It isn't making an error this time. As it turns out, both of our sets of answers work upon being checked by multiplication as I described above. I thought you might like to know that the method I used is perfectly valid because there are no terms with odd degrees of x present in the equation.

Phoenix: That's a really nifty approach I forgot all about!

Answer 3

F(x) = x^4 + x² + 1

To find the zeroes, set F(x) = 0
x^4 + x² + 1 = 0

x^4 can be rewritten as (x²)², so
(x²)² + x² + 1 = 0

Transpose 1
(x²)² + x² = -1

add 1/4
(x²)² + x² + 1/4 = -1 + 1/4

Factor the left side(perfect square)
(x² + 1/2)² = -3/4

Get the square root of both sides
x² + 1/2 = sqrt(-3)/sqrt(4)

Get sqrt
x² + 1/2 = i sqrt3 /2

transpose 1/2
x² = -1/2 ± i sqrt3 /2

Simplify expression
x² = (-1 ± i sqrt3)/2

Get square root
x = sqrt[(-1 ± i sqrt3)/2]

Get both sqrt
x = sqrt(-1 ± i sqrt 3)/sqrt2

Mult sqrt2/sqrt2
x = sqrt(2)sqrt(-1 ± i sqrt 3)/2

Put 2 inside
x = sqrt(-2 ± 2i sqrt 3)/2

Therefore,
x = sqrt(-2 + 2i sqrt 3)/2
x = sqrt(-2 - 2i sqrt 3)/2

Get complex square root of both x's

sqrt(-2 ± 2i sqrt3)/2
= sqrt(-1/2 ± i sqrt 3/2)

the two roots are
= (-1/2 + i sqrt 3/2)^(1/2)
= (-1/2 - i sqrt 3/2)^(1/2)
Use the formula

(x + yi)^(1/n) = r^(1/n)[cis (Θ - 360k)/n]
k = 0,1,...,n - 1

To do that convert both roots into polar form
r(cos Θ + i sin Θ) = r(cis Θ)
Rectangular form: x + yi = z
(a)
-1/2 + i sqrt 3/2
x = -1/2
y = sqrt3/2

y = r sin Θ
x = r cos Θ
y/x = sinΘ/cosΘ
tanΘ = y/x = (sqrt3/2)/(-1/2)
tan Θ = -sqrt3
Θ = -60
r = sqrt(x² + y²)
r = sqrt(1/4 + 3/4) = sqrt(1) = -1
z = -(cis -60)

-1/2 - i sqrt 3/2
tan Θ = y/x = (-sqrt3/2)/(-1/2) = sqrt3
Θ = 60
= -(cis 60)

Now use the formula z^(1/2) for z = -(cis -60) and z = -(cis 60)
for
z = -(cis -60)
z^(1/2) = (-1)^(1/2)(cis (-60)/2)
z^(1/2) = i(cis -30)
z^(1/2) = i(cos -30 + i sin -30)
z^(1/2) = i(sqrt3/2 - 1/2 i)
z^(1/2) = -1/2(i²) + i sqrt3/2
z^(1/2) = 1/2 + i sqrt3/2

z^(1/2) = (-1)^(1/2)(cis (-60 - 360)/2)
z^(1/2) = i(cis (-420)/2)
z^(1/2) = i(cos -210 + i sin -210)
z^(1/2) = i(-sqrt3/2 + 1/2 i)
z^(1/2) = -(i²)sqrt3/2 + 1/2 i
z^(1/2) = -1/2 - i sqrt3/2

for
z = -(cis 60)
z^(1/2) = (-1)^(1/2)(cis (60)/2)
z^(1/2) = i(cis 30)
z^(1/2) = i(cos 30 + i sin 30)
z^(1/2) = i(sqrt3/2 + 1/2 i)
z^(1/2) = 1/2(i²) + i sqrt3/2
z^(1/2) = -1/2 + i sqrt3/2

z^(1/2) = (-1)^(1/2)(cis (60 - 360)/2)
z^(1/2) = i(cis (-300)/2)
z^(1/2) = i(cos -150 + i sin -150)
z^(1/2) = i(-sqrt3/2 - 1/2 i)
z^(1/2) = -(i²)1/2 + sqrt3/2 i
z^(1/2) = 1/2 - i sqrt3/2

Therefore, the roots are

x = 1/2 + i sqrt3/2
x = -1/2 + i sqrt3/2
x = 1/2 - i sqrt3/2
x = -1/2 - i sqrt3/2

^_^

Answer 4

There are no real roots. They are all complex.
The complex roots are
-1/2 - (1/2)sqrt(3)i
-1/2 + (1/2)sqrt(3)i
1/2 - (1/2)sqrt(3)i
1/2 + (1/2)sqrt(3)i
Note: The above method that everyone is using does not work.
I can't recall the method that you need to use, but I do know that the four answers I have given are correct.
Also,
http://www.quickmath.com/
does produce the correct approximate answers. This is where I got my exact answers.

Answer 5

Guess x4 + x2 + 1 = (x2 + ax + 1)((x2 - ax + 1).
Work this out. You find x4 + (2-a2)x2 + 1
Now 2 - a2 must be 1. So a = 1 or -1. It results in
x4 + x2 + 1 = (x2 + x + 1)((x2 - x + 1).
The begin equation can be written as
(x2 + x + 1)((x2 - x + 1) = 0.
It means x2 + x + 1 = 0 or x2 - x + 1 = 0.
Since (+ or - 1)^2 -4*1*1 < 0 there are no solutions with real numbers.
The four complex solutions are
( -1 + 3i)/2 ----- ( -1 - 3i)/2
(+1 + 3i)/2 ----- (+1 - 3i)/2

Answer 6

the zeroes are:

+or- sq root of (0.5+i.5i) and +or- sq root of (0.5-1.5i)
four complex zeroes....



since u have asked zeroes we are then about to find such real or complex numbers which make f(x)=0 hence the roots of the equation

step 1:
substitute x^2 by X and x^4 by X^2

step 2:
you will get a quadratic equation
solve the equation u will get two complex roots they are your X

step 3:
since x^2=X (we substituted earlier) then x has to be a complex number we assume x as a+ib. then square (a+ib) and compare it with the roots from step 2....

step 4
you will be able to find a and b algebraically...

step 5
since there are two roots and each roots have two possibe values + or - of roots .. hence you will have four zeroes....
which matches that any equation the number of roots is equal to the highest degree..

Answer 7

Solve x^4+x^2+1==0.

Let y == x^2.
Then y^2+y+1==0.
So y == (-1 +- sqrt(-3))/2.
So x == +- sqrt((-1 +- isqrt3)/2) (four values).

Answer 8

x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1)

x = (.5)(1 - isqrt(3)), (.5)(1 + isqrt(3)), (-.5)(1 - isqrt(3)), (-.5)(1 + isqrt(3))

Answer 9

i cant