Just Curious

As we all know, if $a_0+a_1x+...+a_nx^n=0$ has a solution of $p+qi$ where $\left\{ a_0,a_1,...,a_n \right\} \subset \mathbb{R}\$ , then $p-qi$ is also a solution.

Similarly, if $a_0+a_1x+...+a_nx^n=0$ has a solution of $p+q\sqrt { m }$ where $\left\{ a_{ 0 },a_{ 1 },...,a_{ n } \right\} \subset \mathbb{Q}\ \quad and\quad \sqrt { m } \in \mathbb{I}\$ , then $p-q\sqrt { m }$ is also a solution.

I was playing around with it until I suddenly got curious. If $a_{ 0 }+a_{ 1 }x+...+a_{ n }x^{ n }=0\quad (n\ge 4\quad and\quad n\in \mathbb{N}\ )$ has a solution of $p\sqrt { m } +qi\quad (\sqrt { m } \in \mathbb{I}\ )$ , then is it safe to assume that it also has the solutions of $p\sqrt { m } -qi,\quad -p\sqrt { m } +qi,\quad -p\sqrt { m } -qi\quad$ ?

I tried it for $n=4$ , and it worked.

Can someone confirm this for all values of n's?

#Algebra #Equation

Easy Math Editor

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Comments

Yes, your observation is correct. Can you figure out how to prove it?

Calvin Lin Staff - 6 years, 7 months ago

Let $p + q\sqrt{m}$ be a root of a polynomial $f(x)$ .

We can write $f(x)$ = $(x-(p + q\sqrt{m})$ $(x-(p - q\sqrt{m}))$ $a(x)$ + $r(x)$

$r(x)$ 's degree would be 1,

$\implies r(x) = ax + b$

If we get $a=b=0$ ,then $p - q\sqrt{m}$ will also be root of $f(x)$

Put $x = p + q\sqrt{m}$ & get $a=b=0$ .

(This is not a proof but a hint.)

Harsh Shrivastava - 6 years, 7 months ago

대입해서 복소수상등 무리수상등을 동시해하시면되는데 그게 힘들죠... 증명해보고잇어요!

찬홍 민 - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$