Irrational polynomial equation

In summary, the conversation discusses the polynomial equation that \sqrt{2} + \sqrt{3} satisfies, with various suggestions and clarifications being made. It is mentioned that there is not one unique polynomial that satisfies this equation, but it can be represented as a linear combination of monomials. The conversation also touches on the idea of using conjugates to create a higher-order polynomial equation. Overall, the focus is on finding a polynomial equation that satisfies certain criteria, such as having integer coefficients.
  • #1
ciel
2
0
what's the polynomial equation which sqrt2 + sqrt3 satisfies ?
 
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  • #2
How about x-sqrt2-sqrt3?
 
  • #3
There is not one unique polynomial that (sprt2 + sqrt3) satisfies. Perhaps you should elaborate a bit as to what terms, degree, etc. that you require in your polynomial.
 
  • #4
And keep in mind that a polynomial is nothing more than a linear combination of monomials...
 
  • #5
What about, [tex]x- \pi[/tex] ? or [tex]x^3 - 31[/tex] =] ?
 
  • #6
ciel said:
what's the polynomial equation which sqrt2 + sqrt3 satisfies ?
If you want to get rid of the roots, then it'd be something like

[tex]x = \sqrt{2}+\sqrt{3}[/tex]
[tex]x^{2} = 2+3+2\sqrt{6}[/tex]
[tex]x^{2}-5 = 2\sqrt{6}[/tex]
[tex](x^{2}-5)^{2} = 24[/tex]

Rearrange and clean up a bit.
 
  • #7
yeah, i seems alright that way. well, i has to be polynomial, but i couldn't just go upto that part, so i was confused. thnx anyway :)
 
  • #8
Gib Z said:
What about, [tex]x- \pi[/tex] ? or [tex]x^3 - 31[/tex] =] ?
:confused:
 
  • #9
Hurkyl: And keep in mind that a polynomial is nothing more than a linear combination of monomials...

From the standpoint of Galois Theory, we can build that form from the conjugates of the two order two equations, which produces an order 4 equation:

The four products of the form: [tex]\prod (X-(\pm\sqrt2\pm\sqrt3)[/tex] = X^4-10X^2+1.
 
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  • #10
It would have been nice if you had posted the actual question. I suspect it was NOT "Find a polynomial equation that [itex]\sqrt{2}+ \sqrt{3}[/itex] satisfies". I suspect rather that it was something like "Find a polynomial equation, with integer coefficients, that [itex]\sqrt{2}+ \sqrt{3}[/itex] satisfies".
 
  • #11
EES said:
:confused:

I'm just joking with you man, try it on yours calculator, [itex]\sqrt{2}+\sqrt{3}[/itex] is a valid approximation of pi to several digits. Similar thing for the other equation.
 

What is an irrational polynomial equation?

An irrational polynomial equation is an equation that contains both rational and irrational coefficients, as well as variables raised to irrational powers. This means that the solutions to the equation may not be expressible as rational numbers, and may involve complex numbers.

How do you solve an irrational polynomial equation?

Solving an irrational polynomial equation involves using algebraic techniques such as factoring, the rational root theorem, and the quadratic formula. In some cases, the equation may need to be approximated numerically using methods such as Newton's method.

What is the degree of an irrational polynomial equation?

The degree of an irrational polynomial equation is determined by the highest power of the variable present in the equation. For example, an equation with a term containing x^4 would be a fourth-degree polynomial equation. The degree of an irrational polynomial equation can also be determined by the number of solutions it has.

What are some real-world applications of irrational polynomial equations?

Irrational polynomial equations are used in various fields of science and engineering, such as in modeling physical systems, predicting population growth, and analyzing financial data. They are also used in computer science for tasks such as image and signal processing.

Can irrational polynomial equations have imaginary solutions?

Yes, irrational polynomial equations can have imaginary solutions. This is because the irrational coefficients and powers in the equation can lead to complex numbers as solutions. Imaginary solutions can be identified when solving the equation using the quadratic formula, as they involve taking the square root of a negative number.

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