# Difference between revisions of "Conjugate Root Theorem"

Hashtagmath (talk | contribs) (→Uses) |
Duck master (talk | contribs) (added proof of Conjugate Root Theorem) |
||

Line 1: | Line 1: | ||

=Theorem= | =Theorem= | ||

− | The Conjugate Root Theorem states that if <math>P(x)</math> is a polynomial with real coefficients, and <math>a+bi</math> is a root of the equation <math>P(x) = 0</math>, where <math>i = \sqrt{-1}</math>, then <math>a-bi</math> is also a root. | + | The Conjugate Root Theorem states that if <math>P(x)</math> is a [[polynomial]] with real [[coefficients]], and <math>a+bi</math> is a [[root]] of the equation <math>P(x) = 0</math>, where <math>i = \sqrt{-1}</math>, then <math>a-bi</math> is also a root. |

A similar theorem states that if <math>P(x)</math> is a polynomial with rational coefficients and <math>a+b\sqrt{c}</math> is a root of the polynomial, then <math>a-b\sqrt{c}</math> is also a root. | A similar theorem states that if <math>P(x)</math> is a polynomial with rational coefficients and <math>a+b\sqrt{c}</math> is a root of the polynomial, then <math>a-b\sqrt{c}</math> is also a root. | ||

+ | |||

+ | ==Proof== | ||

+ | Suppose that <math>P(a + bi) = 0</math>. Then <math>\overline{P(a + bi)} = 0</math>. However, we know that <math>\overline{P(a + bi)} = \overline{P}(\overline{a + bi}) = P(a - bi)</math>, where we define <math>\overline{P}</math> to be the polynomial with the coefficients replaced with their [[complex conjugate|complex conjugates]]; we know that <math>\overline{P} = P</math> by the assumption that <math>P</math> has real coefficients. Thusly, we show that <math>P(a - bi) = 0</math>, and we are done. | ||

==Uses== | ==Uses== | ||

− | This has many uses. If you get a fourth degree polynomial, and you are given that a number in the form of <math>a+bi</math> is a root, then you know that <math>a-bi</math> in the root. Using the [[Factor Theorem]], you know that <math>(x-(a+bi))(x-(a-bi))</math> is also a root. Thus, you can multiply that out, and divide it by the original polynomial, to get a depressed quadratic equation. Of course, it doesn't have to be a fourth degree polynomial. It could just simplify it a bit. | + | This has many uses. If you get a fourth degree polynomial, and you are given that a number in the form of <math>a+bi</math> is a root, then you know that <math>a-bi</math> in the root. Using the [[Factor Theorem]], you know that <math>(x-(a+bi))(x-(a-bi))</math> is also a root. Thus, you can multiply that out, and divide it by the original polynomial, to get a [[quadratic formula|depressed quadratic equation]]. Of course, it doesn't have to be a fourth degree polynomial. It could just simplify it a bit. |

{{stub}} | {{stub}} | ||

[[Category:Theorems]] | [[Category:Theorems]] | ||

+ | [[Category:Algebra]] | ||

+ | [[Category:Polynomials]] |

## Revision as of 03:03, 9 December 2019

# Theorem

The Conjugate Root Theorem states that if is a polynomial with real coefficients, and is a root of the equation , where , then is also a root. A similar theorem states that if is a polynomial with rational coefficients and is a root of the polynomial, then is also a root.

## Proof

Suppose that . Then . However, we know that , where we define to be the polynomial with the coefficients replaced with their complex conjugates; we know that by the assumption that has real coefficients. Thusly, we show that , and we are done.

## Uses

This has many uses. If you get a fourth degree polynomial, and you are given that a number in the form of is a root, then you know that in the root. Using the Factor Theorem, you know that is also a root. Thus, you can multiply that out, and divide it by the original polynomial, to get a depressed quadratic equation. Of course, it doesn't have to be a fourth degree polynomial. It could just simplify it a bit.

*This article is a stub. Help us out by expanding it.*