Polynomial Roots Homework: Find 2nd Polynomial with Squared Roots

In summary, the given cubic with roots \alpha,\beta,\gamma can be used to find a new polynomial with roots \alpha^2,\beta^2,\gamma^2 by taking the sum of the squares of the roots and using the fact that (\alpha\beta +\alpha\gamma+\beta\gamma)^2=\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2. This yields a polynomial in the form a_1^2x^3+(2a_1c_1-b_1^2)x^2+\frac{c_1^2-2b_1d_1}{a_1^2}x+d_1
  • #1
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Homework Statement


Given a general cubic [tex]a_1x^3+b_1x^2+c_1x+d_1=0[/tex] has roots [tex]\alpha,\beta,\gamma[/tex]

find the polynomial [tex]a_2x^3+b_2x^2+c_2x+d_2=0[/tex] that has roots [tex]\alpha ^2,\beta ^2,\gamma ^2[/tex]

The Attempt at a Solution


[tex]\alpha ^2+\beta ^2+\gamma ^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma)[/tex]

Thus, [tex]\alpha ^2+\beta ^2+\gamma ^2=(\frac{-b_1}{a_1})^2-2(\frac{c_1}{a_1})=\frac{b_1^2-2a_1c_1}{a_1^2}[/tex]

Therefore, [tex]-\frac{b_2}{a_2}\equiv -\frac{b_1^2-2a_1c_1}{a_1^2}[/tex]

So the new polynomial is now in the form:

[tex]a_1^2x^3+(2a_1c_1-b_1^2)x^2+c_2x+d_2=0[/tex]


Also, [tex]\alpha ^2\beta ^2\gamma ^2=(\alpha\beta\gamma)^2[/tex]

Thus, [tex]\alpha ^2\beta ^2\gamma ^2=(\frac{d_1}{a_1})^2=\frac{d_1^2}{a_1^2}[/tex]

Therefore, [tex]-\frac{d_2}{a_2}\equiv -\frac{d_1^2}{a_1^2}[/tex]

So now the polynomial is:

[tex]a_1^2x^3+(2a_1c_1-b_1^2)x^2+c_2x+d_1^2=0[/tex]

In order to find [tex]c_2[/tex] in terms of the coefficients of the first polynomial, I'll need to express
[tex]\alpha ^2\beta ^2+\alpha ^2\gamma ^2+\beta ^2\gamma ^2[/tex] in terms of sum of roots one, two and three at a time, using the similar idea as was done to find the sum of the squared roots one at a time. However, I'm unsure how to do this. Please help.
 
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  • #2
You'd need to consider

[tex](\alpha^2 + \beta^2 + \gamma^2)^2[/tex]
 
  • #3
You can also substitute x = sqrt(y) in the equation. Then that immediately guarantees that the roots in terms of y are the squares of the roots of the original equation. To get rid of the fractional powers of y, you bring the y^(1/2) and y^(3/2) terms to one side of the equation and then you square both sides.
 
  • #4
This question was in an exam I had:

Count Iblis said:
You can also substitute x = sqrt(y) in the equation. Then that immediately guarantees that the roots in terms of y are the squares of the roots of the original equation. To get rid of the fractional powers of y, you bring the y^(1/2) and y^(3/2) terms to one side of the equation and then you square both sides.
Yes I had first considered to substitute x=sqrt(y) but after realizing it wasn't a polynomial anymore and from what our teacher said previously in class "if it isn't a polynomial then it won't work". I didn't consider squaring to form it into a 3rd degree polynomial again, basically because my trust in it working wasn't there after what my teacher had said.

rock.freak667 said:
You'd need to consider

[tex](\alpha^2 + \beta^2 + \gamma^2)^2[/tex]
This is the approach I took to find the roots, but I'm unsure what to do with the long expansion:

[tex](\alpha ^2+\beta ^2+\gamma ^2)^2=\alpha ^4+\beta ^4+\gamma ^4-2(\alpha ^2\beta ^2+\alpha ^2\gamma ^2+\beta ^2\gamma ^2)[/tex]

As for the 3 roots to the 4th degree, is it really going to be that complicated to solve this problem in this way?
 
  • #5
With these sorts of problems there is usually always a straightforward brute force method available which often involves solving eqations to get things exactly right. But if you think carefully, you can often find a method that yields the answer without much effort.

See here another such problem:

https://www.physicsforums.com/archive/index.php/t-8259.htmlbr/t-263816.html
 
  • #6
Count Iblis said:
But if you think carefully, you can often find a method that yields the answer without much effort.
If you're referring to the post you made in that thread, indeed your solution wasn't the conventional brute force method, but the problem is that I wouldn't have been able to think carefully in that way.
 
  • #7
Ok, you missed this one:

[tex]\alpha\beta +\alpha\gamma+\beta\gamma=\frac{c_1}{a_1}[/tex]

[tex](\alpha\beta +\alpha\gamma+\beta\gamma)^2=\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2+2(\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\beta\gamma^2)=[/tex]
[tex]=\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2+2\alpha\beta\gamma(\alpha+\beta+\gamma)[/tex]


Can you continue?

Regards.
 
  • #8
Дьявол said:
Ok, you missed this one:

[tex]\alpha\beta +\alpha\gamma+\beta\gamma=\frac{c_1}{a_1}[/tex]

[tex](\alpha\beta +\alpha\gamma+\beta\gamma)^2=\alpha^2\beta^2+\alph a^2\gamma^2+\beta^2\gamma^2+2(\alpha^2\beta\gamma+ \alpha\beta^2\gamma+\alpha\beta\gamma^2)[/tex]
[tex]=\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2+ 2\alpha\beta\gamma(\alpha+\beta+\gamma)[/tex]




Can you continue?

Regards.


Oh jeez that is MUCH more simple! I guess rock.freak667 and I went off and chose a much more complicated expansion, to our doom :biggrin: Thanks Дьявол


ok so basically just to finish it off:

[tex]\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2=(\alpha\beta +\alpha\gamma+\beta\gamma)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)=(\frac{c_1}{a_1})^2-2(\frac{-d_1}{a_1})(\frac{-b_1}{a_1})=\frac{c_1^2-2b_1d_1}{a_1^2}[/tex]

Therefore the polynomial is:

[tex]a_1^2x^3+(2a_1c_1-b_1^2)x^2+\frac{c_1^2-2b_1d_1}{a_1^2}x+d_1^2=0[/tex]
 

1. What is a polynomial?

A polynomial is a mathematical expression with one or more terms, each consisting of a variable raised to a non-negative integer power, and coefficients that are real numbers. Examples of polynomials include x, 2x + 5, and x2 + 3x - 1.

2. What are roots of a polynomial?

The roots of a polynomial are the values of the variable that make the polynomial equal to zero. These values are also called solutions or zeros. For example, the roots of the polynomial x2 - 4 are 2 and -2, since (2)2 - 4 = 0 and (-2)2 - 4 = 0.

3. What does it mean to have squared roots in a polynomial?

A polynomial with squared roots means that the roots of the polynomial are the square of a number. In other words, the roots can be expressed as x2 for some value of x. For example, in the polynomial x2 - 9, the roots are 3 and -3, since (3)2 - 9 = 0 and (-3)2 - 9 = 0.

4. How do I find a 2nd polynomial with squared roots?

To find a 2nd polynomial with squared roots, you can start by writing the polynomial in the form of (x - a)(x - b), where a and b are the roots of the polynomial. For a polynomial with squared roots, the roots will be the square of a number, so you can choose any two numbers that are the square of each other, such as 4 and -4. Then, you can expand the expression to get the polynomial in its standard form.

5. Why is finding polynomials with squared roots useful?

Finding polynomials with squared roots can be useful in solving certain problems in mathematics, physics, and engineering. For example, in physics, polynomials with squared roots can represent motion in a straight line with constant acceleration. In engineering, polynomials with squared roots can represent the response of a system to a given input. Additionally, finding polynomials with squared roots can help in understanding the behavior and properties of higher degree polynomials.

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