Roots of a fourth degree polynomial

In summary, the equation z^4 - z^2 + 1 = 0 can be expressed as the product of two quadratics, (z^2 + a1z + 1)(z^2 + a2z + 1). By equating coefficients, it can be determined that a1 = -a2 and a1a2 = -2. Therefore, the sum of the roots of the equation is √2 and the product of the roots is -1. By using the quadratic formula, it can be shown that the sum of two roots of the equation is √(3 ± √5), which means that the correct answer is (iii) √(3 + √5).
  • #1
V0ODO0CH1LD
278
0

Homework Statement



z^4 - z^2 + 1 = 0 is an equation in ℂ.

Which of the following alternatives is the sum of two roots of this equation:

(i) 2√3; (ii) -(√3)/2; (iii) (√3)/2; (iv) -i; (v) i/2

Homework Equations

The Attempt at a Solution



All I know is that the sum of all roots should equal 0, because that's the coefficient in front of the second highest degree term. And that the sum of the roots squared should equal 2, because that is the coefficient of the second highest degree term plus -2 times the third highest degree term.
 
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  • #2
V0ODO0CH1LD said:

Homework Statement



z^4 - z^2 + 1 = 0 is an equation in ℂ.
The equation is quadratic in form. Let x = z2 and use the quadratic formula to find values of x (= z2).
V0ODO0CH1LD said:
Which of the following alternatives is the sum of two roots of this equation:

(i) 2√3; (ii) -(√3)/2; (iii) (√3)/2; (iv) -i; (v) i/2

Homework Equations




The Attempt at a Solution



All I know is that the sum of all roots should equal 0, because that's the coefficient in front of the second highest degree term.
No, the coefficient of the z2 term is -1, not 0.
V0ODO0CH1LD said:
And that the sum of the roots squared should equal 2, because that is the coefficient of the second highest degree term plus -2 times the third highest degree term.
 
  • #3
V0ODO0CH1LD said:

Homework Statement



z^4 - z^2 + 1 = 0 is an equation in ℂ.

Which of the following alternatives is the sum of two roots of this equation:

(i) 2√3; (ii) -(√3)/2; (iii) (√3)/2; (iv) -i; (v) i/2

Homework Equations




The Attempt at a Solution



All I know is that the sum of all roots should equal 0, because that's the coefficient in front of the second highest degree term. And that the sum of the roots squared should equal 2, because that is the coefficient of the second highest degree term plus -2 times the third highest degree term.

It is a quadratic equation for z2. Solve it, then take the square roots of the solutions.


ehild
 
  • #4
Okay, so I said x = z^2 in z^4 - z^2 + 1 = 0 and solved it as if it were a quadratic:

x^2 - x + 1 = 0 <=> (x - (1/2))^2 + 3/4 = 0

(x - (1/2))^2 = -3/4 <=> x - 1/2 = ±√(-3/4)

x = ±√(3)i/2 + 1/2

Now I can just square the roots of the equation for x and I get the roots of the equations for z?

Wait, ehild said I should take the square root of the roots. Why!?
 
Last edited:
  • #5
V0ODO0CH1LD said:
Okay, so I said x = z^2 in z^4 - z^2 + 1 = 0 and solved it as if it were a quadratic:

x^2 - x + 1 = 0 <=> (x - (1/2))^2 + 3/4 = 0

(x - (1/2))^2 = -3/4 <=> x - 1/2 = ±√(-3/4)

x = ±√(3)i/2 + 1/2

Now I can just square the roots of the equation for x and I get the roots of the equations for z?
Think about what you're doing. If you square x, you'll get z4. You want z.
 
  • #6
Wow, thanks! I feel kinda of stupid now.. Anyway, I got to z = √(±√(3)i/2 + 1/2). Which means the sum of two roots of the equation could equal √(√(3)i/2 + 1/2) + √(-√(3)i/2 + 1/2). Any tips on how to simplify this to a point where it looks like one of the answers in the original question?
 
  • #7
V0ODO0CH1LD said:
Wow, thanks! I feel kinda of stupid now.. Anyway, I got to z = √(±√(3)i/2 + 1/2).

Which means the sum of two roots of the equation could equal √(√(3)i/2 + 1/2) + √(-√(3)i/2 + 1/2).
I don't know what you did to get the above.
V0ODO0CH1LD said:
Any tips on how to simplify this to a point where it looks like one of the answers in the original question?

I think the best way to go is to work with x = 1/2 ± i (√3/2), and write this in polar form, or r(cosθ + i sinθ). Here r = 1, so that simplifies things a bit, and the angles should be pretty easy to calculate. Getting a square root is pretty easy - you just take the square root of the magnitude (r) and divide the angle by 2.

For example, consider the complex number i, which in polar form is 1(cos ##\pi/2## + i sin ##\pi/2##). √i = 1(cos ##\pi/4## + i sin ##\pi/4##) = (√2/2) + (√2/2)i. This is one of the square roots of i. You can verify this by multiplying (√2/2) + (√2/2)i by itself.
 
  • #8
V0ODO0CH1LD said:

Homework Statement



z^4 - z^2 + 1 = 0 is an equation in ℂ.

Which of the following alternatives is the sum of two roots of this equation:

(i) 2√3; (ii) -(√3)/2; (iii) (√3)/2; (iv) -i; (v) i/2

Homework Equations

The Attempt at a Solution



All I know is that the sum of all roots should equal 0, because that's the coefficient in front of the second highest degree term. And that the sum of the roots squared should equal 2, because that is the coefficient of the second highest degree term plus -2 times the third highest degree term.

Rather than actually calculating the roots, a much quicker way to do this would be to use the "sum of roots" and "product of roots" formulae for a quadratic. Let two roots of the quartic be [itex]z_1[/itex] and [itex]z_2[/itex].

Then you know that:

[tex]z_1^2 + z_2^2 = 1[/tex]
[tex]z_1^2z_2^2 = 1[/tex]

Now use those equations and figure out what values [itex]{(z_1 + z_2)}^2[/itex] can take, and take the square root of that. There's only one choice that matches.
 
  • #9
The quartic, z4 - z2 + 1, can be expressed as the product of two quadratics. Because of symmetry we can suppose that they are of the following form, where a and b are real coefficients.

[itex]\displaystyle z^4-z^2+1[/itex]
[itex]\displaystyle =(z^2+az+1)(z^2+bz+1)[/itex]

[itex]\displaystyle =z^4+(a+b)z^3+(2+ab)z^2+(a+b)z+1[/itex]​

Equating coefficients of powers of z gives a+b=0 and 2+ab=-1.

These can be solved for a & b.
 
  • #10
V0ODO0CH1LD said:
Wow, thanks! I feel kinda of stupid now.. Anyway, I got to z = √(±√(3)i/2 + 1/2). Which means the sum of two roots of the equation could equal √(√(3)i/2 + 1/2) + √(-√(3)i/2 + 1/2). Any tips on how to simplify this to a point where it looks like one of the answers in the original question?
The existing posts, noting that you don't actually need to find all the roots, are the right way to go. But fwiw, the easiest way to spot the simplification is to look at it geometrically. Where in the plane is 1 + i√3? What is that in (r, θ)?
 
  • #11
SammyS said:
The quartic, z4 - z2 + 1, can be expressed as the product of two quadratics. Because of symmetry we can suppose that they are of the following form, where a and b are real coefficients.

[itex]\displaystyle z^4-z^2+1[/itex]
[itex]\displaystyle =(z^2+az+1)(z^2+bz+1)[/itex]

[itex]\displaystyle =z^4+(a+b)z^3+(2+ab)z^2+(a+b)z+1[/itex]​

Equating coefficients of powers of z gives a+b=0 and 2+ab=-1.

These can be solved for a & b.

If I solve for a and b on those I would get +√3 and -√3 which add up to 0 which is not one of the answers.
Curious3141 said:
Rather than actually calculating the roots, a much quicker way to do this would be to use the "sum of roots" and "product of roots" formulae for a quadratic. Let two roots of the quartic be [itex]z_1[/itex] and [itex]z_2[/itex].

Then you know that:

[tex]z_1^2 + z_2^2 = 1[/tex]
[tex]z_1^2z_2^2 = 1[/tex]

Now use those equations and figure out what values [itex]{(z_1 + z_2)}^2[/itex] can take, and take the square root of that. There's only one choice that matches.

I don't get the last thing. You mean to find the roots of (z^2)^2 - z^2 + 1 = 0 as if they were the roots of a quadratic and then take the square root of those?

EDIT: sorry, I meant find the roots of an equation like x^2 - x + 1 = 0 and take the square root of those!
 
  • #12
SammyS said:
The quartic, z4 - z2 + 1, can be expressed as the product of two quadratics. Because of symmetry we can suppose that they are of the following form, where a and b are real coefficients.

[itex]\displaystyle z^4-z^2+1[/itex]
[itex]\displaystyle =(z^2+az+1)(z^2+bz+1)[/itex]

[itex]\displaystyle =z^4+(a+b)z^3+(2+ab)z^2+(a+b)z+1[/itex]​

Equating coefficients of powers of z gives a+b=0 and 2+ab=-1.

These can be solved for a & b.
V0ODO0CH1LD said:
If I solve for a and b on those I would get +√3 and -√3 which add up to 0 which is not one of the answers.
[itex]\displaystyle \sqrt{3}\ \ \text{ and }\ -\sqrt{3}\ [/itex] are not the roots of anything in this problem. They're the coefficients of z in the respective quadratics which factor the given quartic in this problem.

So, find the roots of each of those quadratics:
[itex]\displaystyle (z^2+\sqrt{3}z+1)\ \ \text{ and }\ (z^2-\sqrt{3}z+1)\ .[/itex]​
This ia not as elegant a solution as those offered by others, but it does work.

Once you get the answer this way, go back and see how the suggestions of others also lead you to the answer.
 
  • #13
V0ODO0CH1LD said:
Wow, thanks! I feel kinda of stupid now.. Anyway, I got to z = √(±√(3)i/2 + 1/2). Which means the sum of two roots of the equation could equal √(√(3)i/2 + 1/2) + √(-√(3)i/2 + 1/2). Any tips on how to simplify this to a point where it looks like one of the answers in the original question?

It is correct, the roots are the square roots of either [itex]1/2+i\sqrt{3}/2[/itex] or [itex]1/2-i\sqrt{3}/2[/itex] . You know that a complex number can be written in the form r(cos(θ)+isin(θ)). What are r and theta?
You have learned something about the roots of a complex number. What was it?

ehild
 
  • #14
Yeah, the roots are:

+ cos (π/6) + sin(π/6)i;
+ cos (π/6) – sin(π/6)i;
– cos (π/6) + sin(π/6)i;
– cos (π/6) – sin(π/6)i;

Which when added give ±√3 or ±i.

But according to what the others have said, I didn't need to actually find the roots in the first place. I can see how the other ways (Curious3141's, SammyS', Mark44's) to get the roots would work. I just don't see how I could have used those properties of quadratics to find the sums.
 
  • #15
Curious' method gives the sum of roots directly.

From the quadratic equation for z2 you get that z12+z22=1 and z12z22=1 (Vieta's formulas). The second equation means that z1z2=±1.
(z1+z2)2=z12+z22+2z1z2=1±2.

ehild
 
  • #16
V0ODO0CH1LD said:
Yeah, the roots are:

+ cos (π/6) + sin(π/6)i;
+ cos (π/6) – sin(π/6)i;
– cos (π/6) + sin(π/6)i;
– cos (π/6) – sin(π/6)i;

Which when added give ±√3 or ±i.

But according to what the others have said, I didn't need to actually find the roots in the first place. I can see how the other ways (Curious3141's, SammyS', Mark44's) to get the roots would work. I just don't see how I could have used those properties of quadratics to find the sums.
Yes, this is correct. (Well, the sum can also be zero.)

As for my method, it did involve finding all of the roots. As follows:

[itex]\displaystyle z^4-z^2+1[/itex]
[itex]\displaystyle =(z^2+\sqrt{3}z+1)(z^2-\sqrt{3}z+1)=0[/itex]​
So that [itex]\displaystyle \ z^2+\sqrt{3}z+1=0\ \ \text{ or }\ z^2-\sqrt{3}z+1=0\ .[/itex]

Using the quadratic formula to solve those gives:
[itex]\displaystyle z=\frac{-\sqrt{3}\pm i }{2}\ , \ \ z=\frac{\sqrt{3}\pm i }{2}\ .[/itex]​
 

1. What is a fourth degree polynomial?

A fourth degree polynomial is a mathematical expression that contains a variable raised to the fourth power, along with other terms such as constants and coefficients. It is also known as a quartic polynomial.

2. How do you find the roots of a fourth degree polynomial?

To find the roots of a fourth degree polynomial, you can use a variety of methods such as factoring, the quadratic formula, or the rational root theorem. There is no one specific method that works for all fourth degree polynomials, so it is important to explore different approaches.

3. Can a fourth degree polynomial have complex roots?

Yes, a fourth degree polynomial can have complex roots. This means that the solutions to the polynomial are not real numbers, but instead involve imaginary numbers. Complex roots often come in conjugate pairs, where one root is the complex conjugate of the other.

4. How many roots does a fourth degree polynomial have?

A fourth degree polynomial can have a maximum of four roots, but it is possible for some of those roots to be repeated. For example, a fourth degree polynomial with two distinct roots and one repeated root would have a total of three roots.

5. What is the relationship between the number of roots and the degree of a polynomial?

The fundamental theorem of algebra states that a polynomial of degree n will have exactly n complex roots. This means that a fourth degree polynomial can have a maximum of four roots, but it is possible for some of those roots to be complex numbers.

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