1. ## Contradiction to Irrational Roots Come in Conjugate Pairs?

During class, we discussed a polynomial while learning about the upper bound theorem and Descarte's Rule of Signs:
2x^3 + x^2 - 8x - 3

Upon applying many theorems to analyze this polynomial and graphing it as well, we discovered that there are two negative real roots and one positive real root.

Using the rational root theorem, we discovered that none of the three real roots are rational because they don't match any of the 1,3 / 1,2 possible rational roots from the rational root theorem.

However, this means there are three irrational roots. That's the problem, we can't figure out why this is possible, because mathematical theorems say that irrational roots come in conjugate pairs.

Can anybody explain why this is possible, or whether there is some flaw in the irrational root theorem? Thanks.

2. Originally Posted by XanthanPower
During class, we discussed a polynomial while learning about the upper bound theorem and Descarte's Rule of Signs:
2x^3 + x^2 - 8x - 3

Upon applying many theorems to analyze this polynomial and graphing it as well, we discovered that there are two negative real roots and one positive real root.

Using the rational root theorem, we discovered that none of the three real roots are rational because they don't match any of the 1,3 / 1,2 possible rational roots from the rational root theorem.

However, this means there are three irrational roots. That's the problem, we can't figure out why this is possible, because mathematical theorems say that irrational roots come in conjugate pairs.

Can anybody explain why this is possible, or whether there is some flaw in the irrational root theorem? Thanks.
Hi XanthanPower,

I think you're mixing up "irrational" with "complex". Complex roots come in conjugate pairs, not irrational roots.

3. Originally Posted by XanthanPower
During class, we discussed a polynomial while learning about the upper bound theorem and Descarte's Rule of Signs:
2x^3 + x^2 - 8x - 3

Upon applying many theorems to analyze this polynomial and graphing it as well, we discovered that there are two negative real roots and one positive real root.

Using the rational root theorem, we discovered that none of the three real roots are rational because they don't match any of the 1,3 / 1,2 possible rational roots from the rational root theorem.

However, this means there are three irrational roots. That's the problem, we can't figure out why this is possible, because mathematical theorems say that irrational roots come in conjugate pairs.

Can anybody explain why this is possible, or whether there is some flaw in the irrational root theorem? Thanks.
To quote from

http://mathforum.org/library/drmath/view/52607.html:

"The irrational conjugate roots theorem says:

"Let p(x) be any polynomial with rational coefficients. If
a + b*sqrt(c) is a root of p(x), where sqrt(c) is irrational and
a and b are rational, then another root is a - b*sqrt(c).

"The problem is that you have no guarantee that the root of a generic
cubic can be written in the form x = a + b*sqrt(c). "

4. What reason do you have to believe that any root is of the form $a+ b\sqrt{c}$? Those are not the only types of irrational numbers.

5. Originally Posted by HallsofIvy
What reason do you have to believe that any root is of the form $a+ b\sqrt{c}$? Those are not the only types of irrational numbers.
Yes, that is exactly the problem with attempting to apply the irrational conjugate roots theorem to the cubic equation in the O.P. (as I, or rather Dr. Math, said).

6. ## Re: Contradiction to Irrational Roots Come in Conjugate Pairs?

The roots of cubic equations often involve cube roots, rather than square roots. (And often there are square roots inside the cube roots.) So the irrational conjugate root theorem does not apply.

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### do irrational roots of any polynomial with real coefficients always occur in pairs??

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