Difference between revisions of "Conjugate Root Theorem"

(Uses)
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=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.
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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.
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==Proof==
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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.
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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]]
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[[Category:Algebra]]
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[[Category:Polynomials]]

Revision as of 03:03, 9 December 2019

Theorem

The Conjugate Root Theorem states that if $P(x)$ is a polynomial with real coefficients, and $a+bi$ is a root of the equation $P(x) = 0$, where $i = \sqrt{-1}$, then $a-bi$ is also a root. A similar theorem states that if $P(x)$ is a polynomial with rational coefficients and $a+b\sqrt{c}$ is a root of the polynomial, then $a-b\sqrt{c}$ is also a root.

Proof

Suppose that $P(a + bi) = 0$. Then $\overline{P(a + bi)} = 0$. However, we know that $\overline{P(a + bi)} = \overline{P}(\overline{a + bi}) = P(a - bi)$, where we define $\overline{P}$ to be the polynomial with the coefficients replaced with their complex conjugates; we know that $\overline{P} = P$ by the assumption that $P$ has real coefficients. Thusly, we show that $P(a - bi) = 0$, 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 $a+bi$ is a root, then you know that $a-bi$ in the root. Using the Factor Theorem, you know that $(x-(a+bi))(x-(a-bi))$ 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.

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